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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