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.

---

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:

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

---

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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