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
$\small{\cos 2x \cos 6x~=~\frac{\cos 8x \,+\,\cos 4x}{2}}$
•
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.
The link below gives a few more miscellaneous examples:
Exercise 23.3
In the next section, we will see a few more solved examples.
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