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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