23.7 - Solved Examples on Integration Using Trigonometric Identities

In the previous section, we saw the application of trigonometric identities in the process of integration. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.9
Find the following integrals:
(i) $\small{\int{\left[\sin 4x \sin 8x \right]dx}}$

(ii) $\small{\int{\left[\sin x \sin 2x \sin 3x \right]dx}}$

(iii) $\small{\int{\left[\cos 2x \cos 4x \cos 6x \right]dx}}$

Solution:
Part (i):
1. We have the identity:
$\small{\cos A\,-\,\cos B\,=\,-2 \sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{-(\cos A\,-\,\cos B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{8x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{4x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 12x and B = 4x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 4x \sin 8x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-(\cos 12x\,-\,\cos 4x)}{2}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos 4x \,-\,\cos 12x}{2}\right] \, dx}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\cos 4x \right] \, dx}~-~\frac{1}{2} \int{\left[\cos 12x \right] \, dx}}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (4x) is 4.
• So we put u = 4x
⇒ $\small{\frac{du}{dx}~=~4}$
⇒ 4 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{4 \cos 4x}{4} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{4} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{4}\,+\,C_1~=~ \frac{\sin 4x}{8}\,+\,C_1}$

Second term:
For this term also, we use the method of substitution.
(i) The derivative of (12x) is 12.
• So we put u = 12x
⇒ $\small{\frac{du}{dx}~=~12}$
⇒ 12 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{12 \cos 12x}{12} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{12} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{12}\,+\,C_2~=~ \frac{\sin 12x}{24}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 4x \sin 8x \right]dx}}    & {~=~}    &{\frac{\sin 4x}{8}\,+\,C_1~-~\frac{\sin 12x}{24}\,-\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin 4x}{8}\,-\,\frac{\sin 12x}{24}\,+\,C}    \\ \end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii): $\small{\int{\left[\sin x \sin 2x \sin 3x \right]dx}}$
• The given function can be rearranged into another form. We will do the rearrangement in three stages.

Stage I:
• x is odd, 2x is even and 3x is odd.
• We must do only two operations:
   ♦ add odd to odd
   ♦ add even to even
• Only then we will be able to divide by 2
• So we will choose (sin x sin 3x) for stage I

1. We have the identity:
$\small{\cos A\,-\,\cos B\,=\,-2 \sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\sin \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{-(\cos A\,-\,\cos B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{3x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{1x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 4x and B = 2x

• So (sin x sin 3x) can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin x \sin 3x}    & {~=~}    &{\frac{-(\cos 4x\,-\,\cos 2x)}{2}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos 2x \,-\,\cos 4x}{2}}    \\ \end{array}}$

• Now the given function can be rearranged as:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
• In stage II, we will rearrange the first term (sin 2x cos 2x)

Stage II: sin 2x cos 2x

1. We have the identity:
$\small{\sin 2A\,=\,2 \sin A \cos A}$
• From this, we get: $\small{\sin A \cos A\,=\,\frac{\sin 2A}{2}}$

2. So for our present problem, we can write: A = 2x
• So (sin 2x cos 2x) can be rearranged as: $\small{\frac{\sin 4x}{2}}$

• Now the given function can be rearranged as:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\sin 2x\cos 4x}{2} }$
• In stage III, we will rearrange (sin 2x cos 4x)

Stage III: (sin 2x cos 4x)

1. We have the identity:
$\small{\sin A\,-\,\sin B\,=\,-2 \cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B)}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 6x and B = 2x

• So (sin 2x cos 4x) can be rearranged as:
$\small{\frac{\sin 6x \,-\, \sin 2x}{2}}$

• Now the given function can be rearranged in the final form:
sin x sin 2x sin 3x = (sin x sin 3x) sin 2x
= $\small{\left(\frac{\cos 2x\,-\,\cos 4x}{2} \right)\sin 2x}$
= $\small{\frac{\sin 2x \cos 2x\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\sin 2x\cos 4x}{2} }$
= $\small{\frac{\frac{\sin 4x}{2}\,-\,\left(\frac{\sin 6x \,-\, \sin 2x}{2} \right)}{2} }$
= $\small{\frac{\sin 4x}{4}\,-\,\frac{\sin 6x}{4}\,+\,\frac{\sin 2x}{4} }$

• Now we can begin the integration process.
• Based on the solved examples that we discussed so far, we can write two formulas:

(i) $\small{\int{\left[\sin mx \right]dx}~=~\frac{-\cos mx}{m}}$

(ii) $\small{\int{\left[\cos mx \right]dx}~=~\frac{\sin mx}{m}}$

• The above two formulas can be used directly while solving problems. So we can easily write the integral:
$\small{\int{\left[\frac{\sin 4x}{4}\,-\,\frac{\sin 6x}{4}\,+\,\frac{\sin 2x}{4} \right]dx}}$

$\small{~=~\left[\frac{-\cos 4x}{4(4)}\,+\,C_1 \right]\,-\,\left[\frac{-\cos 6x}{4(6)}\,+\,C_2 \right]\,+\,\left[\frac{-\cos 2x}{4(2)}\,+\,C_3 \right]}$

$\small{~=~\frac{1}{4}\left[\frac{-\cos 4x}{4}\,+\,\frac{\cos 6x}{6}\,-\,\frac{\cos 2x}{2}\right]\,+\,C}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii): $\small{\int{\left[\cos 2x \cos 4x \cos 6x \right]dx}}$
• The given function can be rearranged into another form. We will do the rearrangement in three stages.

Stage I:
• 2x is even, 4x is even and 6x is even.
• We must do only two operations:
   ♦ add odd to odd
   ♦ add even to even
• Only then we will be able to divide by 2
• Here all terms are even. So we can choose in any order we like. We will choose (cos 2x cos 4x) for stage I

1. We have the identity:
$\small{\cos A\,+\,\cos B\,=\,2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right) \,=\,\frac{\cos A\,+\,\cos B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 6x and B = 2x

• So (cos 2x cos 4x) can be rearranged as:
$\small{\cos 2x \cos 4x~=~\frac{\cos 6x \,+\,\cos 2x}{2}}$

• Now the given function can be rearranged as:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
• In stage II, we will rearrange the first term (cos²6x)

Stage II: cos²6x

1. We have the identity: $\small{\cos 2A \,=\, 2 \cos^2 A \,-\, 1}$
• From this, we get: $\small{\cos^2 A \,=\,\frac{1\,+\,\cos 2A}{2}}$

2. So for our present problem, we can write:
$\small{\cos^2 6 x \,=\,\frac{1\,+\,\cos 12x}{2}}$

• Now the given function can be rearranged as:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\cos 2x\cos 6x}{2} }$
• In stage III, we will rearrange (cos 2x cos 6x)

Stage III: (cos 2x cos 6x)

1. We have the identity:
$\small{\cos A\,+\,\cos B\,=\,2 \cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos \left(\frac{A+B}{2} \right) \cos \left(\frac{A - B}{2} \right) \,=\,\frac{\cos A\,+\,\cos B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{6x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 8x and B = 4x

• So (cos 2x cos 6x) can be rearranged as:
$\small{\cos 2x \cos 6x~=~\frac{\cos 8x \,+\,\cos 4x}{2}}$

• Now the given function can be rearranged in the final form:
cos 2x cos 4x cos 6x = (cos 2x cos 4x) cos 6x
= $\small{\left(\frac{\cos 6x\,+\,\cos 2x}{2} \right)\cos 6x}$
= $\small{\frac{\cos^2 6x\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\cos 2x\cos 6x}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x}{2}\,+\,\frac{\cos 8x \,+\,\cos 4x}{2}}{2} }$
= $\small{\frac{\frac{1\,+\,\cos 12x\,+\,\cos 8x\,+\,\cos 4x}{2}}{2} }$
= $\small{\frac{1}{4}\,+\,\frac{\cos 12x}{4}\,+\,\frac{\cos 8x}{4}\,+\,\frac{\cos 4x}{4}}$

• Now we can begin the integration process.
• Based on the solved examples that we discussed so far, we can write two formulas:

(i) $\small{\int{\left[\sin mx \right]dx}~=~\frac{-\cos mx}{m}}$

(ii) $\small{\int{\left[\cos mx \right]dx}~=~\frac{\sin mx}{m}}$

• The above two formulas can be used directly while solving problems. So we can easily write the integral:
$\small{\int{\left[\frac{1}{4}\,+\,\frac{\cos 12x}{4}\,+\,\frac{\cos 8x}{4}\,+\,\frac{\cos 4x}{4} \right]dx}}$

$\small{~=~\left[\frac{x}{4}\,+\,C_1 \right]\,+\,\left[\frac{\sin 12x}{4(12)}\,+\,C_2 \right]\,+\,\left[\frac{\sin 8x}{4(8)}\,+\,C_3 \right]\,+\,\left[\frac{\sin 4x}{4(4)}\,+\,C_4 \right]}$

$\small{~=~\frac{1}{4}\left[x\,+\,\frac{\sin 12x}{12}\,+\,\frac{\sin 8x}{8}\,+\,\frac{\sin 4x}{4}\right]\,+\,C}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

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The link below gives a few more miscellaneous examples:

Exercise 23.3

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In the next section, we will see a few more solved examples.

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23.6 - Integration Using Trigonometric Identities

In the previous section, we saw some solved examples demonstrating integration by substitution. In this section, we will see the application of trigonometric identities in the process of integration. We have already seen some examples in the previous section. For example, in the solved example 23.7(ii) of the previous section, we used the identity:
$\small{\sin A \,-\, \sin B ~=~2 \cos \left(\frac{A+B}{2} \right) \sin \left(\frac{A-B}{2} \right)}$

Now we will see some advanced problems

Solved example 23.8
Find the following integrals:
(i) $\small{\int{\left[\cos^2 x \right]dx}}$

(ii) $\small{\int{\left[\sin 2x \cos 3x \right]dx}}$

(iii) $\small{\int{\left[\sin^3 x \right]dx}}$

(iv) $\small{\int{\left[\sin 3x \cos 4x \right]dx}}$

Solution:
Part (i):
1. We have the identity: $\small{\cos 2A \,=\, 2 \cos^2 A \,-\, 1}$
• From this, we get: $\small{\cos^2 A \,=\,\frac{1\,+\,\cos 2A}{2}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\cos^2 x \right]dx}}    & {~=~}    &{\int{\left[\frac{1\,+\,\cos 2x}{2}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[1 \right] \, dx}~+~\frac{1}{2} \int{\left[\cos 2x \right] \, dx}}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[1 \right] \, dx~=~\frac{x}{2}\,+\,C_1}}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (2x) is 2.
• So we put u = 2x
⇒ $\small{\frac{du}{dx}~=~2}$
⇒ 2 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{2 \cos 2x}{2} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\cos u}{2} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{\sin u}{2}\,+\,C_2~=~ \frac{\sin 2x}{4}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\cos^2 x\right]dx}}    & {~=~}    &{\frac{x}{2}\,+\,C_1~+~\frac{\sin 2x}{4}\,+\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{\sin 2x}{4}\,+\,C}    \\ \end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii):
1. We have the identity:
$\small{\sin A\,-\,\sin B\,=\,2 \cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right)}$
• From this, we get: $\small{\cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{3x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{2x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 5x and B = 1x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 2x \cos 3x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin 5x\,-\,\sin x}{2}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\sin 5x \right] \, dx}~-~\frac{1}{2} \int{\left[\sin x \right] \, dx}}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (5x) is 5.
• So we put u = 5x
⇒ $\small{\frac{du}{dx}~=~5}$
⇒ 5 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{5 \sin 5x}{5} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\sin u}{5} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{-\cos u}{5}\,+\,C_1~=~ \frac{-\cos 5x}{10}\,+\,C_1}$

Second term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[\sin x \right] \, dx~=~\frac{-\cos x}{2}\,+\,C_2}}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 2x \cos 3x \right]dx}}    & {~=~}    &{\frac{-\cos 5x}{10}\,+\,C_1~-~\frac{-\cos x}{2}\,+\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos 5x}{10}\,+\,\frac{\cos x}{2}\,+\,C}    \\ \end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\sin^3 x}    & {~=~}    &{\sin x \,\sin^2 x}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin x(1\,-\,\cos^2 x)}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sin x \,-\,\sin x \, \cos^2 x}    \\ \end{array}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[ \sin^3 x \right]dx}}    & {~=~}    &{\int{\left[\sin x \,-\,\sin x \, \cos^2 x \right]dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sin x \right]}dx~-~\int{\left[\sin x \, \cos^2 x \right]}dx}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\int{\left[\sin x \right] }\, dx~=~-\cos x \,+\,C_1}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (cos x) is −sin x.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}~=~-\sin x}$
⇒ (−sin x)dx = du

(ii) So we want:
$\small{\int{\left[(-1)(-1) \sin x \, \cos^2 x \right]dx}~=~ \int{\left[(-1) u^2 \right]du}}$

• This integration gives:
$\small{(-1) \frac{u^3}{3}\,+\,C_2~=~ (-1)\frac{\cos^3 x}{3}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x\right]dx}}    & {~=~}    &{-\cos x \,+\,C_1~-~ (-1)\frac{\cos^3 x}{3}\,+\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{-\cos x ~+~ \frac{\cos^3 x}{3}\,+\,\,C}    \\ \end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Alternate method:

1. We have the identity: $\small{\sin 3A\,=\,3 \sin A \,-\,4 \sin^3 A}$
• From this, we get: $\small{\sin^3 A \,=\,\frac{3 \sin A \,-\,\sin 3A}{4}}$

2. So for our present problem, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x \right]dx}}    & {~=~}    &{\int{\left[\frac{3 \sin x \,-\,\sin 3x}{4}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{3}{4} \int{\left[\sin x \right] \, dx}~-~\frac{1}{4} \int{\left[\sin 3x \right] \, dx}}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
• This term is easy. We can directly write:
$\small{\frac{3}{4} \int{\left[\sin x \right] \, dx~=~\frac{-3 \cos x}{4}\,+\,C_1}}$

Second term:
For this term, we use the method of substitution.
(i) The derivative of (3x) is 3.
• So we put u = 3x
⇒ $\small{\frac{du}{dx}~=~3}$
⇒ 3 dx = du

(ii) So we want:
$\small{\frac{1}{4} \int{\left[\frac{3 \sin 3x}{3} \right]dx}~=~\frac{1}{4} \int{\left[\frac{\sin u}{3} \right]du}}$

• This integration gives:
$\small{\frac{1}{4} \frac{-\cos u}{3}\,+\,C_2~=~ \frac{-\cos 3x}{12}\,+\,C_2}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x\right]dx}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,C_1~-~\frac{-\cos 3x}{12}\,+\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos 3x}{12}\,+\,C}    \\ \end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

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We see that, the results obtained by the two methods are different. But their equality can be proved by using trigonometric identities. This is shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos 3x}{12}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{4 \cos^3 x\,-\,3 \cos x}{12}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{4 \cos^3 x}{12}\,-\,\frac{3 \cos x}{12}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-3 \cos x}{4}\,+\,\frac{\cos^3 x}{3}\,-\,\frac{\cos x}{4}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{-4 \cos x}{4}\,+\,\frac{\cos^3 x}{3}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-\cos x\,+\,\frac{\cos^3 x}{3}}    \\ \end{array}}$

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Part (iv):
1. We have the identity:
$\small{\sin A \,-\,\sin B\,=\,2 \cos \left(\frac{A+B}{2} \right)\,\sin \left(\frac{A-B}{2} \right)}$
• From this, we get: $\small{\cos\left(\frac{A+B}{2} \right) \sin \left(\frac{A - B}{2} \right) \,=\,\frac{\sin A\,-\,\sin B}{2}}$

2. So for our present problem, we can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{A+B}{2}}    & {~=~}    &{4x}    \\ {~\color{magenta}    2    }    &{{}}    &{{\frac{A-B}{2}}}    & {~=~}    &{3x}    \\ \end{array}}$

• Solving the two equations, we get:
A = 7x and B = 1x

• So the given function can be rearranged as:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 3x \cos 4x\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin 7x\,-\,\sin x}{2}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[\sin 7x \right] \, dx}~-~\frac{1}{2} \int{\left[\sin x \right] \, dx}}    \\ \end{array}}$

3. The R.H.S has two terms. We will consider each term separately.
First term:
For this term, we use the method of substitution.
(i) The derivative of (7x) is 7.
• So we put u = 7x
⇒ $\small{\frac{du}{dx}~=~7}$
⇒ 7 dx = du

(ii) So we want:
$\small{\frac{1}{2} \int{\left[\frac{7 \sin 7x}{7} \right]dx}~=~\frac{1}{2} \int{\left[\frac{\sin u}{7} \right]du}}$

• This integration gives:
$\small{\frac{1}{2} \frac{-\cos u}{5}\,+\,C_1~=~ \frac{-\cos 7x}{14}\,+\,C_1}$

Second term:
• This term is easy. We can directly write:
$\small{\frac{1}{2} \int{\left[\sin x \right] \, dx~=~\frac{-\cos x}{2}\,+\,C_2}}$

4. Now, based on step 2, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin 3x \cos 4x \right]dx}}    & {~=~}    &{\frac{-\cos 7x}{14}\,+\,C_1~-~\frac{-\cos x}{2}\,+\,C_2}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos 7x}{14}\,+\,\frac{\cos x}{2}\,+\,C}    \\ \end{array}}$                           

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

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In the next section, we will see a few more solved examples.

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23.5 - Solved Examples on Integration by Substitution

In the previous section, we saw some standard integrals of trigonometric functions. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.7
Find the following integrals:
(i) $\small{\int{\left[(4x+2) \sqrt{x^2 + x + 1} \right]dx}}$

(ii) $\small{\int{\left[\frac{1}{1\,-\, \tan x} \right]dx}}$

(iii) $\small{\int{\left[\frac{1}{1\,+\, \cot x} \right]dx}}$

(iv) $\small{\int{\left[\frac{1}{x\,+\, x \log x} \right]dx}}$

Solution:
Part (i):
1. The derivative of (x²+x+1) is 2x+1.
• So we put u = x²+x+1
⇒ $\small{\frac{du}{dx}~=~2x\,+\,1}$
⇒ (2x+1)dx = du

2. So we want:
$\small{\int{\left[(4x+2) \sqrt{x^2 + x + 1} \right]dx}~=~\int{\left[2 \sqrt{u} \right]du}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[2 \sqrt{u}\right] \, du}}    & {~=~}    &{2 \left[\frac{u^{3/2}}{3/2}~+~C_1 \right]~=~\frac{4 u^{3/2}}{3}}~+~C_2    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{4(x^2\,+\,x\,+\,1)^{3/2}}{3}~+~C}    \\ \end{array}}$

• Note that, the constants C₁, C₂ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (ii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,-\,\tan x}}    & {~=~}    &{\frac{1}{1\,-\,(\sin x / \cos x)}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\cos x\,-\,\sin x}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin (\pi/2 \,-\, x)\,-\,\sin x}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{2 \cos(\pi/4) \sin(\pi/4\,-\, x)}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sqrt{2} \sin(\pi/4 \,-\, x)}}    \\ \end{array}}$                           

• Derivative of (π/4 − x) w.r.t x is −1.
• So we put u = (π/4 −x)
⇒ $\small{\frac{du}{dx}~=~-1}$
⇒ −dx = du
• Also, since u = (π/4 −x), we get: x = π/4 − u
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{(-1)(-1)\cos x}{\sqrt{2} \sin(\pi/4 \,-\, x)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-\cos (\pi/4 \,-\, u)}{\sqrt{2} \sin u}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-\cos (\pi/4) \cos u ~-~\sin (\pi/4) \sin u}{\sqrt{2} \sin u}\right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cos u ~-~(1/\sqrt 2)\sin u }{\sqrt{2} \sin u}\right] \, du}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cot u~-~(1/\sqrt 2) }{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2)(\cot u ~+~1)}{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1~+~\cot u)}{2}\right] \, du}}    \\ \end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{-1~-~\cot u}{2}\right] \, du}}    & {~=~}    &{\int{\left[\frac{-1}{2}\right] \, du}~-~\int{\left[\frac{\cot u}{2}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \int{\left[1 \right] \, du}~-~\frac{1}{2} \int{\left[\cot u \right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \left[u\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sin u |\,+\, C_2 \right]}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\frac{1}{2} \left[(\pi/4)\,-\,x\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sin ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,-\,C_1 \right]~+~\frac{1}{2} \left[-\log |\sin ((\pi/4)\,-\,x) |\,-\, C_2 \right]}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log |\sin ((\pi/4)\,-\,x) |\,+\, C_4 \right]}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|\sin (\pi/4) \, \cos x~-~\cos (\pi/4)  \,\sin x \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2) \, \cos x ~-~  (1/\sqrt 2) \,\sin x \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2)(\cos x ~-~\sin x)  \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log (1/\sqrt 2)~-~ \log \left|(\cos x ~-~\sin x)  \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(\cos x ~-~\sin x)  \right |\,+\, C_5 \right]}    \\ {~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{C_3}{2}~-~\frac{\log \left|(\cos x ~-~\sin x)  \right |}{2}\,+\,\frac{C_5}{2}}    \\ {~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,-\,\frac{\log \left|(\cos x ~-~\sin x)  \right |}{2}\,+\,C}    \\ \end{array}}$                            
                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,+\,\cot x}}    & {~=~}    &{\frac{1}{1\,+\,(\cos x / \sin x)}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\sin x\,+\,\cos x}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\cos (\pi/2 \,-\, x)\,+\,\cos x}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{2 \cos(\pi/4) \cos(\pi/4\,-\, x)}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\sin x}{\sqrt{2} \cos(\pi/4 \,-\, x)}}    \\ \end{array}}$                           

• Derivative of (π/4 −x) w.r.t x is −1.
• So we put u = (π/4 −x)
⇒ $\small{\frac{du}{dx}~=~-1}$
⇒ −dx = du
• Also, since u = (π/4 −x), we get: x = π/4 − u
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{(-1)(-1)\sin x}{\sqrt{2} \cos(\pi/4 \,-\, x)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{-\sin (\pi/4 \,-\, u)}{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-\sin (\pi/4) \cos u ~+~\cos (\pi/4) \sin u}{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) \cos u ~+~(1/\sqrt 2)\sin u }{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2) ~+~(1/\sqrt 2) \tan u }{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1/\sqrt 2)(1~-~\tan u)}{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{-(1~-~\tan u)}{2}\right] \, du}}    \\ \end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{-1~+~\tan u}{2}\right] \, du}}    & {~=~}    &{\int{\left[\frac{-1}{2}\right] \, du}~+~\int{\left[\frac{\tan u}{2}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \int{\left[1 \right] \, du}~+~\frac{1}{2} \int{\left[\tan u \right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{-1}{2} \left[u\,+\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos u |\,+\, C_2 \right]}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\frac{1}{2} \left[(\pi/4)\,-\,x\,+\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,-\,C_1 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log |\cos ((\pi/4)\,-\,x) |\,+\, C_2 \right]}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|\cos (\pi/4) \, \cos x~+~\sin (\pi/4)  \,\sin x \right |\,+\, C_2 \right]}    \\ {~\color{magenta}    8    }    &{{}}   &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2) \, \cos x ~+~  (1/\sqrt 2) \,\sin x \right |\,+\, C_2 \right]}    \\ {~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(1/\sqrt 2)(\cos x ~+~\sin x)  \right |\,+\, C_2 \right]}    \\ {~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log (1/\sqrt 2)~-~ \log \left|(\cos x ~-~\sin x)  \right |\,+\, C_2 \right]}    \\ {~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[-\log \left|(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,+\,\frac{C_3}{2}~-~\frac{\log \left|(\cos x ~+~\sin x)  \right |}{2}\,+\,\frac{C_4}{2}}    \\ {~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2}\,-\,\frac{\log \left|(\cos x ~+~\sin x)  \right |}{2}\,+\,C}    \\ \end{array}}$                           
                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

Part (iv):
1. The given expression can be rearranged as follows:
$\small{\frac{1}{x\,+\, x \log x}~=~\frac{1}{x(1\,+\, \log x)}}$                     

• Derivative of (1+log x) w.r.t x is: (0+1/x) = 1/x.
• So we put u = 1+ log x
⇒ $\small{\frac{du}{dx}~=~\frac{1}{x}}$
⇒ $\small{du~=~\frac{1}{x} {dx}}$

2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{x\,+\, x \log x}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{1}{x(1\,+\, \log x)}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1}{u}\right] \, du}}    \\ \end{array}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1}{u}\right] \, du}}    & {~=~}    &{\int{\left[\frac{1}{x(1\,+\, \log x)}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1}{u}\right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\log|(1\,+\, \log x)|~+~C}    \\ \end{array}}$

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The link below gives a few more solved examples:

Exercise 23.2

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In the next section, we will see a few more solved examples.

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23.4 - Standard Integrals of Trigonometric Functions

In the previous section, we saw integration by substitution. We saw some solved examples also. In this section, we will see some standard integrals of trigonometric functions.

In the table of integrals that we saw at the beginning of this chapter, there is the integral of sin x and also the integral of cos x. But altogether, there are six trigonometric functions. They are sine, cos, tan, csc, sec and cot. So there are four results missing from the table. This is because, integrals of sine and cosine can be obtained easily by inspection. The integrals of other four need to be derived using appropriate methods. We can use the method of substitution.

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First we will find $\small{\int{\left[\tan x \right]dx}}$
1. We have: $\small{\tan x ~=~\frac{\sin x}{\cos x}}$
• −sin x is the derivative of cos x
• So we put u = cos x
⇒ $\small{\frac{du}{dx}}$ = −sin x
⇒ −sin x dx = du

2. So we want:
$\small{\int{\left[\tan x \right]dx}}$
= $\small{\int{\left[\frac{\sin x}{\cos x} \right]dx}~=~\int{\left[\frac{(-1) \sin x}{(-1) \cos x} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1) \cos x} \right]du}~=~\int{\left[\frac{1}{(-1) u} \right]du}~=~(-1)\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{- \log |u| \,+\,C~=~-\log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~\log\left[\frac{1}{|u|}  \right]\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\tan x \right]dx~=~\log\left[\frac{1}{|\cos x|}  \right]\,+\,C~=~\log \left[|\sec x| \right]\,+\,C}}$

• We can write:
$\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$

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Next we will find $\small{\int{\left[\cot x \right]dx}}$
1. We have: $\small{\cot x ~=~\frac{\cos x}{\sin x}}$
• cos x is the derivative of sin x
• So we put u = sin x
⇒ $\small{\frac{du}{dx}}$ = cos x
⇒ cos x dx = du

2. So we want:
$\small{\int{\left[\cot x \right]dx}}$
= $\small{\int{\left[\frac{\cos x}{\sin x} \right]dx}}$
= $\small{\int{\left[\frac{1}{\sin x} \right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{\log[|u|] \,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$

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Next we will find $\small{\int{\left[\sec x \right]dx}}$
1. We have: $\small{\sec x ~=~\frac{\sec x(\sec x\,+\,\tan x)}{\sec x\,+\,\tan x}~=~\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x}}$
• (sec² x + sec x tan x) is the derivative of (tan x + sec x)
• So we put u = tan x + sec x
⇒ $\small{\frac{du}{dx}~=~\sec^2 x\,+\, \sec x \, \tan x}$
⇒ (sec² x + sec x tan x) dx = du

2. So we want:
$\small{\int{\left[\sec x \right]dx}}$
= $\small{\int{\left[\frac{\sec^2 x\,+\, \sec x \, \tan x}{\sec x\,+\,\tan x} \right]dx}}$
= $\small{\int{\left[\frac{1}{\sec x\,+\,\tan x} \right]du}~=~\int{\left[\frac{1}{ u} \right]du}}$

• This integration gives:
$\small{\log[|u|] \,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$

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Finally we will find $\small{\int{\left[\csc x \right]dx}}$
1. We have: $\small{\csc x ~=~\frac{\csc x(\csc x\,+\,\cot x)}{\csc x\,+\,\cot x}~=~\frac{\csc^2 x\,+\, \csc x \, \cot x}{\csc x\,+\,\cot x}}$
• (−csc² x − csc x cot x) is the derivative of (cot x + csc x)
• So we put u = cot x + csc x
⇒ $\small{\frac{du}{dx}~=~-\csc^2 x\,-\, \csc x \, \cot x}$
⇒ −(csc² x + csc x cot x) dx = du

2. So we want:
$\small{\int{\left[\csc x \right]dx}}$
= $\small{\int{\left[\frac{(-1)(\csc^2 x\,+\, \csc x \, \cot x)}{(-1)(\csc x\,+\,\cot x)} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1)(\csc x\,+\,\cot x)} \right]du}~=~(-1) \int{\left[\frac{1}{u} \right]du}}$

• This integration gives:
$\small{(-1) \log[|u|] \,+\,C~=~\log[|u|]^{-1} \,+\,C~=~ \log[\frac{1}{|u|}] \,+\,C}$

3. Substituting for u, we get:
$\small{\log \left[\left|\frac{1}{\csc x\,+\,\cot x} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{(\csc x\,+\,\cot x)(\csc x\,-\,\cot x)} \right| \right]\,+\,C~=~\log \left[\left|\frac{\csc x\,-\,\cot x}{\csc^2 x\,-\,\cot^2 x} \right| \right]\,+\,C}$

= $\small{\log \left[\left|\frac{\csc x\,-\,\cot x}{1} \right| \right]\,+\,C~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$ 

---

Let us compile the six results related to trigonometric ratios:
1. $\small{\int{\left[\sin x \right]dx~=~- \cos x \,+\,C}}$

2. $\small{\int{\left[\cos x \right]dx~=~ \sin x \,+\,C}}$

3. $\small{\int{\left[\tan x \right]dx~=~\log \left[|\sec x| \right]\,+\,C}}$

4. $\small{\int{\left[\cot x \right]dx~=~\log \left[|\sin x| \right]\,+\,C}}$

5. $\small{\int{\left[\sec x \right]dx~=~\log \left[|\sec x\,+\,\tan x| \right]\,+\,C}}$

6. $\small{\int{\left[\csc x \right]dx}~=~\log\left[|\csc x\,-\,\cot x| \right]\,+\,C}$

---

Now we will see some solved examples:

Solved example 23.6
Find the following integrals:
(i) $\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}}$

(ii) $\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}}$

(iii) $\small{\int{\left[\frac{1}{1\,+\, \tan x} \right]dx}}$

Solution:
Part (i):
1. −sin x is the derivative of cos x. There is no −ve sign in the question. But it can be adjusted by multiplying two −ve signs.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}~=~-\sin x}$
⇒ −sinx dx = du
• Also, when u = cos x, we get: sin²x = 1 − u²
2. So we want:
$\small{\int{\left[\sin^3 x \, \cos^2 x \right]dx}~=~\int{\left[(-1)(-1) \sin^3 x \, \cos^2 x \right]dx}}$

= $\small{\int{\left[(-1)\sin^2 x \,\, u^2 \right]du}~=~\int{\left[(-1)(1 - u^2) \, u^2 \right]du}}$
• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[(-1)(1 - u^2) \, u^2\right] \, du}}    & {~=~}    &{\int{\left[-u^2 + u^4 \, \right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[-u^2 \, \right] \, du}~+~\int{\left[u^4 \, \right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{- u^{3}}{3}\,+\,C_1 \right]~+~\left[\frac{u^{5}}{5} \,+\,C_2 \right]}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{- u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C}    \\ \end{array}}$                           


3. Substituting for u, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin^3 x \, \cos^2 x\right] \, du}}    & {~=~}    &{\frac{-u^{3}}{3}\,+\, \frac{u^{5}}{5} \,+\, C}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{-\cos^{3} x}{3}\,+\, \frac{\cos^{5} x}{5} \,+\, C}    \\ \end{array}}$

Part (ii):
1. The derivative of (x+a) is 1.
• So we put u = x+a
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
• Also, since u = x+a, we get: x = u−a
2. So we want:
$\small{\int{\left[\frac{\sin x}{\sin(x+a)} \right]dx}~=~\int{\left[\frac{\sin x}{\sin(u)} \right]dx}}$

= $\small{\int{\left[\frac{\sin (u-a)}{\sin(u)} \right]du}}$

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{\sin x}{\sin (x+a)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\sin (u-a)}{\sin u}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\sin u \cos a \,-\, \cos u \sin a }{\sin u}\right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\cos a \,-\, \cot u \sin a\right] \, du}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\cos a \right] \, du}~-~\int{\left[\cot u \sin a\right] \, du}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\cos a \int{\left[1 \right] \, du}~-~ \sin a \int{\left[\cot u \right] \, du}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\cos a \left[u \,+\,C_1 \right]~-~\sin a \left[\log |\sin u | \,+\,C_2 \right]}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{(\cos a)u \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin u | \,-\,(\sin a) C_2}    \\ {~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{(\cos a)(x+a) \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2}    \\ {~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,+\, a \cos a \,+\,(\cos a)C_1 ~-~(\sin a)\log |\sin (x+a) | \,-\,(\sin a) C_2}    \\ {~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,+\, C_3 \,+\,C_4 ~-~(\sin a)\log |\sin (x+a) | \,-\,C_5}    \\ {~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{x \cos a \,-\, (\sin a)\log |\sin (x+a) | \,+\,C}    \\ \end{array}}$

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.                            

Part (iii):
1. The given expression can be rearranged as follows:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1}{1\,+\,\tan x}}    & {~=~}    &{\frac{1}{1\,+\,(\sin x / \cos x)}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin x\,+\,\cos x}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sin x\,+\,\sin(\pi/2 \,-\, x)}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{2 \sin(\pi/4) \cos(x\,-\, \pi/4)}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}}    \\ \end{array}}$                           

• Derivative of (x−π/4) w.r.t x is 1.
• So we put u = (x−π/4)
⇒ $\small{\frac{du}{dx}~=~1}$
⇒ dx = du
• Also, since u = (x−π/4), we get: x = u + π/4
2. So we want:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{\cos x}{\sqrt{2} \cos(x\,-\, \pi/4)}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{\cos (u\,+\, \pi/4)}{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos u \cos (π/4)~-~\sin u \sin(π/4)}{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{\cos u (1/\sqrt 2)~-~\sin u (1/\sqrt 2)}{\sqrt{2} \cos u}\right] \, du}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{(1/\sqrt 2)~-~\tan u (1/\sqrt 2)}{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{(1/\sqrt 2)(1~-~\tan u)}{\sqrt{2}}\right] \, du}}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{1~-~\tan u}{2}\right] \, du}}    \\ \end{array}}$                           

• This integration can be done as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1~-~\tan u}{2}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{1}{2}\right] \, du}~-~\int{\left[\frac{\tan u}{2}\right] \, du}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \int{\left[1 \right] \, du}~-~\frac{1}{2} \int{\left[\tan u \right] \, du}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[u\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec u |\,+\, C_2 \right]}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,-\,(\pi/4)\,+\,C_1 \right]~-~\frac{1}{2} \left[\log |\sec (x\,-\,(\pi/4)) |\,+\, C_2 \right]}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log |\sec (x\,-\,(\pi/4)) |\,-\, C_2 \right]}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[- \log \left|\frac{1}{\cos (x\,-\,(\pi/4))} \right |\,-\, C_2 \right]}    \\ {~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\left(\frac{1}{\cos (x\,-\,(\pi/4))}\right)^{-1} \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos (x\,-\,(\pi/4)) \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    9    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, \cos(\pi/4) ~+~\sin x  \,\sin(\pi/4) \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {10}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|\cos x \, (1/\sqrt 2) ~+~\sin x  \,(1/\sqrt 2) \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {11}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(1/\sqrt 2)(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {12}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log (1/\sqrt 2)~+~ \log \left|(\cos x ~+~\sin x)  \right |\,+\, C_4 \right]}    \\ {~\color{magenta}    {13}    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left[x\,+\,C_3 \right]~+~\frac{1}{2} \left[\log \left|(\cos x ~+~\sin x)  \right |\,+\, C_5 \right]}    \\ {~\color{magenta}    {14}    }    &{{}}    &{{}}    & {~=~}    &{\frac{x}{2} \,+\, \frac{1}{2} \left[\log \left|(\cos x ~+~\sin x)  \right | \right]\,+\, C}    \\ \end{array}}$                           

• Note that, the constants C₁, C₂, C₃ etc., can be combined into a single constant C because, all constants, when differentiated, will give zero only.

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In the next section, we will see a few more solved examples.

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23.3 - Methods of Integration

In the previous section, we completed a discussion on integration by method of inspection. This method is not applicable in all cases. So we need to learn about some other methods also. Three prominent methods are:
    ♦ Integration by substitution
    ♦ Integration using partial fractions
    ♦ Integration by parts
• In this section, we will see integration by substitution.

• Recall the chain rule that we used for differentiating functions of the form f(g(x)).
For example:
$\small{\frac{d}{dx}\left[\sin (2x+3)  \right]~=~\cos(2x+3) \times \frac{d}{dx} \left[2x+3 \right]~=~2 \cos(2x+3)}$
• So if the integrand is 2 cos(2x+3), we can use a reverse process to get an integral sin(2x+3)

The reverse process can be explained in 6 steps:
1. We want: $\int{\left[2 \cos (2x+3) \right] dx}$  
2. The integrand is 2 cos(2x+3). We closely examine the integrand and find two parts such that, one of them is the derivative of the other.
• In this case, the two parts are: “2” and “(2x + 3)”
    ♦ “2” is the derivative of “(2x + 3)”
3. We put u = 2x+3
Then $\small{\frac{du}{dx}}$ = 2, which gives 2 dx = du
4. So the expression in (1) can be rewritten as:
$\small{\int{\left[\cos (u) \right] du}}$
• Note that,
    ♦ 2x+3 is replaced by u
    ♦ 2dx is replaced by du
5. Integration of the expression in (4) gives: sin u + C
• Replacing u, we get: sin(2x+3) + C
6. This method is called integration by substitution.
Note that, we did the substitution of (2x+3) by u.

Another example can be written in 6 steps:
1. We want: $\small{\int{\left[6x(3x^2 \,+\, 4)^4 \right] dx}}$  
2. The integrand is 6x (3x² + 4)⁴. We closely examine the integrand and find two parts such that, one of them is the derivative of the other.
• In this case, the two parts are: “6x” and “(3x² + 4)”
    ♦ “6x” is the derivative of “(3x² + 4)”
3. We put u = 3x² + 4
Then $\small{\frac{du}{dx}}$ = 6x, which gives 6x dx = du
4. So the expression in (1) can be rewritten as:
$\int{\left[u^4 \right] du}$  
• Note that,
    ♦ 3x² + 4 is replaced by u
    ♦ 6x dx is replaced by du
5. Integration of the expression in (4) gives: $\small{\frac{u^5}{5} \,+\, C}$
Replacing u, we get: $\small{\frac{(3x^2 \,+\, 4)^5}{5} \,+\, C}$
6. So we used integration by substitution.
Note that, we did the substitution of (3x² + 4) by u.

---

Now we will see some solved examples

Solved example 23.5
Use substitution to find:
(i) $\small{\int{\left[3 x^2 (x^3 \,-\,3)^2 \right]dx}}$

(ii) $\small{\int{\left[\sin(mx) \right]dx}}$

(iii) $\small{\int{\left[\frac{\tan^4 \sqrt{x} \, \sec^2 \sqrt{x}}{\sqrt{x}} \right]dx}}$

(iv) $\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}}$

(v) $\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}}$

(vi) $\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right]dx}}$

Solution:
Part (i):
1. 3x² is the derivative of (x³ - 3)
• So we put u = (x³ - 3)
⇒ $\small{\frac{du}{dx}}$ = 3x²
⇒ 3x² dx = du

2. So we want:
$\small{\int{\left[(u)^2 \right]du}}$
• This integration gives $\small{\frac{u^3}{3}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[3 x^2 (x^3 \,-\,3)^2 \right]dx~=~\int{\left[u^2 \right]du}~=~\frac{u^3}{3}\,+\,C~=~\frac{(x^3 \,-\, 3)^3}{3}\,+\,C}}$

Part (ii):
1. m is the derivative of mx. There is no independent m in the question. But since m is a constant, there will not be any issue.
• So we put u = mx
⇒ $\small{\frac{du}{dx}}$ = m
⇒ m dx = du

2. So we want:
$\small{\int{\left[\frac{m \sin(mx)}{m} \right]dx}}$
= $\small{\int{\left[\frac{\sin(u)}{m} \right]du}}$
• This integration gives $\small{\frac{- \cos(u)}{m}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\sin(mx) \right]dx}~=~\frac{- \cos(mx)}{m }\,+\,C}$

Part (iii):
1. $\small{\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}}}$ is the derivative of $\small{\tan \sqrt{x}}$. There is no independent '2' in the question. But since '2' is a constant, there will not be any issue.
• So we put u = $\small{\tan \sqrt{x}}$
⇒ $\small{\frac{du}{dx}}~=~\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}}$

⇒ $\small{\left(\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}} \right)dx~=~du}$

2. So we want:
$\small{\int{\left[2 \tan^4 \sqrt{x} \,\left(\frac{\sec^2 \sqrt{x}}{2 \sqrt{x}} \right) \right]dx}~=~\int{\left[2\, u^4 \right]du}}$
• This integration gives $\small{\frac{2\,u^5}{5}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\tan^4 \sqrt{x} \, \sec^2 \sqrt{x}}{\sqrt{x}} \right]dx}~=~\frac{2\, \tan^5 \sqrt{x}}{5}\,+\,C}$

Part (iv):
1. $\small{\frac{1}{1\,+\,x^2}}$ is the derivative of $\small{\tan^{-1} x}$.
• So we put u = $\small{\tan^{-1} x}$
⇒ $\small{\frac{du}{dx}~=~\frac{1}{1\,+\,x^2}}$

⇒ $\small{\left(\frac{1}{1\,+\,x^2} \right)dx~=~du}$

2. So we want:
$\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}~=~\int{\left[\sin(u) \right]du}}$
• This integration gives $\small{- \cos(u)\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\sin(\tan^{-1} x)}{1\,+\,x^2} \right]dx}~=~- \cos(\tan^{-1} x)\,+\,C}$

Part (v):
1. −sin x is the derivative of cos x. There is no −ve sign in the question. But since (−1) is a constant, there will not be any issue.
• So we put u = cos x
⇒ $\small{\frac{du}{dx}}$ = −sin x
⇒ −sin x dx = du

2. So we want:
$\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}}$
= $\small{\int{\left[\frac{(-1) \sin x}{(-1) \cos^3 x} \right]dx}}$
= $\small{\int{\left[\frac{1}{(-1) u^3} \right]du}}$

• This integration gives $\small{(-1) \frac{1}{(-2) u^2}\,+\,C~=~ \frac{1}{2 u^2}\,+\,C}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{\sin x}{\cos^3 x} \right]dx}~=~ \frac{1}{2 \cos^2 x}\,+\,C}$

Part (vi):
1. The derivative of (x − 1) is 1. The number '1' will be already available.
• So we put u = x − 1
⇒ $\small{\frac{du}{dx}}$ = 1
⇒ (1) dx = du
• Also, u = x − 1 gives: x = u + 1

2. So we want:
$\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right](1)dx}}$
= $\small{\int{\left[\frac{u\,+\,1}{\sqrt u} \right]du}}$

• This integration can be done as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{u\,+\,1}{\sqrt u}\right] \, dx}}    & {~=~}    &{\int{\left[\frac{u\,+\,1}{\sqrt u} \right] \, dx}~=~\int{\left[\frac{u}{\sqrt u} ~+~\frac{1}{\sqrt u} \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{u^{1/2}}{1} ~+~\frac{u^{-1/2}}{1} \right] \, dx}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\frac{u^{1/2}}{1}  \right] \, dx}~+~\int{\left[\frac{u^{-1/2}}{1} \right] \, dx}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{u^{3/2}}{3/2}\,+\,C_1 \right]~+~\left[\frac{u^{1/2}}{1/2} \,+\,C_2 \right]}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\frac{u^{3/2}}{3/2}\,+\, \frac{u^{1/2}}{1/2} \,+\, C}    \\ {~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 u^{3/2}}{3}\,+\, \frac{2 u^{1/2}}{1} \,+\, C}    \\ \end{array}$

3. Substituting for u, we get:
$\small{\int{\left[\frac{x}{\sqrt{x\,-\,1}} \right]dx}~=~ \frac{2 (x\,-\,1)^{3/2}}{3}\,+\, \frac{2 (x\,-\,1)^{1/2}}{1} \,+\, C}$

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In the next section, we will see some standard integrals of trigonometric functions.

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23.2 - Solved Examples on Method by Inspection

In the previous section, we saw integration by method of inspection. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 23.2
Find the following integrals:
$(i)~\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~~~~(ii)~\int  {\left[x^{2/3} \,+\,1\right]dx}$

$(iii)~\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x}\right] \, dx}~~~~(iv)~\int{\left[\frac{4}{1+x^2}\right] \, dx}$

Solution:
Part (i):
1. We want $\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}$
• This can be written as:
$\int{\left[x~-~\frac{1}{x^2} \right] \, dx}$
⇒ $\int{\left[x~-~x^{-2} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}$
• We search for F and find that:
    ♦ x, is the derivative of $\frac{x^2}{2}$
    ♦ x⁻², is the derivative of $\frac{x^{-1}}{-1}$ which is −(1/x)
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}}    & {~=~}    &{\int{\left[x \right]\,dx}~-~\int{\left[x^{-2} \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^2}{2}\,+\,C_1 \right] ~-~\left[\frac{-1}{x}\,+\,C_2 \right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C_1 \,-\,C_2}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{x^2}{2}\,+\, \frac{1}{x} \,+\,C}    \\ \end{array}$

• This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[\frac{x^3 - 1}{x^2} \right] \, dx}~=~\frac{x^2}{2}\,+\, \frac{1}{x}$ is an anti derivative of $\frac{x^3 - 1}{x^2}$. 

Part (ii):
1. We want $\int{\left[x^{2/3} \,+\,1 \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}$
• We search for F and find that:
    ♦ x^2/3, is the derivative of $\frac{x^{5/3}}{5/3}$
    ♦ 1, is the derivative of x
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[x^{2/3} \,+\,1 \right] \, dx}}    & {~=~}    &{\int{\left[x^{2/3} \right]\,dx}~+~\int{\left[1 \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^{5/3}}{5/3}\,+\,C_1 \right] ~+~\left[x\,+\,C_2 \right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C_1 \,+\,C_2}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{3 x^{5/3}}{5}\,+\, x \,+\,C}    \\ \end{array}$                           

• This is the general form of the required anti derivative.
◼ Remarks:
4 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[x^{2/3} \,+\,1 \right] \, dx}~=~\frac{3 x^{5/3}}{5}\,+\, x$ is an anti derivative of $x^{2/3} \,+\,1$.

Part (iii):
1. We want $\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}$
• By applying property I, this can be written as:
$\int{\left[x^{3/2} \right]\,dx}~+~\int{\left[2 e^x \right]}~-~\int{\left[\frac{1}{x} \right] \, dx}$
• We search for F and find that:
    ♦ x^3/2, is the derivative of $\frac{x^{5/2}}{5/2}$
    ♦ 2e^x, is the derivative of 2e^x
    ♦ 1/x, is the derivative of log|x|
2. So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}}    & {~=~}    &{\int{\left[x^{3/2} \right] \, dx}~+~\int{\left[2 e^x  \right] \, dx}~-~\int{\left[\frac{1}{x} \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{x^{5/2}}{5/2}\,+\,C_1 \right]~+~\left[2 e^x \,+\,C_2 \right]~-~\left[\log|x| \,+\, C_3\right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C_1 \,+\,C_2 \,-\,C_3}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x| \,+\,C}    \\ \end{array}$                           

• This is the general form of the required anti derivative.
◼ Remarks:
3 (magenta color): Addition/subtraction of any two real numbers C₁ and C₂ will give another real number which can be denoted in general as C.

• Or we can write:
$\int{\left[x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x} \right] \, dx}~=~\frac{2 x^{5/2}}{5}\,+\,2 e^x \,-\,\log|x|$ is an anti derivative of $x^{3/2} \,+\, 2 e^x \,-\, \frac{1}{x}$.

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Once we understand the basics, there is no need to write all the detailed steps. So from now on wards, we shall write only the necessary steps.

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Solved example 23.3
Find the following integrals:
$(i)~\int{\left[\sin x \,+\, \cos x \right] \, dx}~~~~~~~(ii)~\int  {\left[\csc x (\csc x \,+\, \cot x)\right]dx}~~~~~~~(iii)~\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\sin x \,+\, \cos x \right] \, dx}}    & {~=~}    &{\int{\left[\sin x  \right] \, dx}~+~\int{\left[\cos x \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[- \cos x\,+\,C_1 \right]~+~\left[\sin x \,+\,C_2 \right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{-\cos x \,+\, \sin x \,+\, C}    \\ \end{array}$

Part (ii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\csc x (\csc x \,+\, \cot x) \right] \, dx}}    & {~=~}    &{\int{\left[\csc^2 x \,+\, \csc x \cot x \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\csc^2 x \right] \, dx}~+~\int{\left[\csc x \cot x \right] \, dx}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[-\cot x \,+\, C_1 \right]~+~\left[-\csc x \,+\,C_2 \right]}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{-\cot x \,-\, \csc x \,+\,C}    \\ \end{array}$

Part (iii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{1 \,-\, \sin x}{\cos^2 x} \right] \, dx}}    & {~=~}    &{\int{\left[\frac{1 }{\cos^2 x} ~-~\frac{\sin x}{\cos^2 x}\right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sec^2 x ~-~ \sec x \tan x \right] \, dx}}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\int{\left[\sec^2 x \right] \, dx}~-~\int{\left[\sec x \tan x \right] \, dx}}    \\ {~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left[\tan x \,+\, C_1 \right]~-~\left[\sec x \,+\,C_2 \right]}    \\ {~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\tan x \,-\, \sec x \,+\,C}    \\ \end{array}$

Part (iv):
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[\frac{4}{1+x^2} \right] \, dx}}    & {~=~}    &{4 \int{\left[\frac{1}{1+x^2} \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{4 \left[\tan^{-1} x \,+\, C \right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4 \tan^{-1} x \,+\, C}    \\ \end{array}$                           
 

Solved example 23.4
Find the anti derivative F of f defined by f (x) = 4x³ − 6, where F(0) = 3
Solution:
1. First we will write the general form of the anti derivative:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\int{\left[4x^3 \,-\,6 \right] \, dx}}    & {~=~}    &{\int{\left[4 x^3 \right] \, dx}~-~\int{\left[6 \right] \, dx}}    \\ {~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\frac{4 x^4}{4} \,+\, C_1 \right]~-~\left[6x \,+\,C_2 \right]}    \\ {~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{x^4 \,-\, 6x \,+\,C}    \\ \end{array}$

• So the general form of the anti derivative is: x⁴ − 6x + C

2.Given that, when x = 0, the anti derivative becomes 3.
• So we can write: x⁴ − 6x + C = 3
⇒ (0)⁴ − 6(0) + C = 3
⇒ C = 3

3. So based on the general form, we can write:
The exact anti derivative is: x⁴ − 6x + 3

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The link below gives a few more solved examples:

Exercise 23.1

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In the next section, we will see Methods of Integration.

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