Calculus/Integration techniques/Trigonometric Integrals

When the integrand is primarily or exclusively based on trigonometric functions, the following techniques are useful.

Powers of Sine and Cosine

We will give a general method to solve generally integrands of the form $\cos ^{m}(x)\cdot \sin ^{n}(x)$ . First let us work through an example.

$\int \cos ^{3}(x)\sin ^{2}(x)dx$

Notice that the integrand contains an odd power of cos. So rewrite it as

$\int \cos ^{2}(x)\sin ^{2}(x)\cos(x)dx$

We can solve this by making the substitution $u=\sin(x)$ so $du=\cos(x)dx$ . Then we can write the whole integrand in terms of $u$ by using the identity

$\cos ^{2}(x)=1-\sin ^{2}(x)=1-u^{2}$ .

So

$\int \cos ^{3}(x)\sin ^{2}(x)dx$	$=\int \cos ^{2}(x)\sin ^{2}(x)\cos(x)dx$
	$=\int (1-u^{2})u^{2}du$
	$=\int u^{2}du-\int u^{4}du$
	$={\frac {u^{3}}{3}}+{\frac {u^{5}}{5}}+C$
	$={\frac {\sin ^{3}(x)}{3}}-{\frac {\sin ^{5}(x)}{5}}+C$

This method works whenever there is an odd power of sine or cosine.

To evaluate $\int \cos ^{m}(x)\sin ^{n}(x)dx$ when either $m$ or $n$ is odd.

• If $m$ is odd substitute $u=\sin(x)$ and use the identity $\cos ^{2}(x)=1-\sin ^{2}(x)=1-u^{2}$ .
• If $n$ is odd substitute $u=\cos(x)$ and use the identity $\sin ^{2}(x)=1-\cos ^{2}(x)=1-u^{2}$ .

Example

Find $\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{3}(x)dx$ .

As there is an odd power of $\sin$ we let $u=\cos(x)$ so $du=-\sin(x)dx$ . Notice that when $x=0$ we have $u=\cos(0)=1$ and when $x={\frac {\pi }{2}}$ we have $u=\cos({\tfrac {\pi }{2}})=0$ .

$\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{3}(x)dx$	$=\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{2}(x)\sin(x)dx$
	$=-\int \limits _{1}^{0}u^{40}(1-u^{2})du$
	$=\int \limits _{0}^{1}u^{40}(2-u^{2})du$
	$=\int \limits _{0}^{5}(u^{40}-u^{42})du$
	$=\left({\frac {u^{41}}{41}}-{\frac {u^{43}}{43}}\right){\Bigg |}_{0}^{1}$
	$={\frac {1}{41}}-{\frac {1}{43}}$

When both $m$ and $n$ are even, things get a little more complicated.

To evaluate $\int \cos ^{m}(x)\sin ^{n}(x)dx$ when both $m$ and $n$ are even.

Use the identities $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$ and $\cos ^{2}(x)={\frac {1+\cos(2x)}{2}}$ .

Example

Find $\int \sin ^{2}(x)\cos ^{4}(x)dx$ .

As $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$ and $\cos ^{2}(x)={\frac {1+\cos(2x)}{2}}$ we have

$\int \sin ^{2}(x)\cos ^{4}(x)dx=\int \left({\frac {1-\cos(2x)}{2}}\right)\left({\frac {1+\cos(2x)}{2}}\right)^{2}dx$

and expanding, the integrand becomes

${\frac {1}{8}}\int \left(1-\cos ^{2}(2x)+\cos(2x)-\cos ^{3}(2x)\right)dx$

Using the multiple angle identities

$I$	$={\frac {1}{8}}\left(\int 1dx-\int \cos ^{2}(2x)dx+\int \cos(2x)dx-\int \cos ^{3}(2x)dx\right)$
	$={\frac {1}{8}}\left(x-{\frac {1}{2}}\int {\Big (}1+\cos(4x){\Big )}dx+{\frac {\sin(2x)}{2}}-\int \cos ^{2}(2x)\cos(2x)dx\right)$
	TODO: CORRECT FORMULA$={\frac {1}{164}}\left(x+\sin(2x)+\int \cos(4x)dx-2\int {\Big (}1-\sin ^{2}(2x){\Big )}\cos(2x)dx\right)$

then we obtain on evaluating

$I={\frac {x}{16}}-{\frac {\sin(4x)}{64}}+{\frac {\sin ^{3}(2x)}{48}}+C$

Powers of Tan and Secant

To evaluate $\int \tan ^{m}(x)\sec ^{n}(x)dx$ .

1. If $n$ is even and $n\geq 2$ then substitute $u=tan(x)$ and use the identity $\sec ^{2}(x)=1+\tan ^{2}(x)$ .
2. If $n$ and $m$ are both odd then substitute $u=\sec(x)$ and use the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$ .
3. If $n$ is odd and $m$ is even then use the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$ and apply a reduction formula to integrate $\sec ^{j}(x)dx$ , using the examples below to integrate when $j=1,2$ .

Example 1

Find $\int \sec ^{2}(x)dx$ .

There is an even power of $\sec(x)$ . Substituting $u=\tan(x)$ gives $du=\sec ^{2}(x)dx$ so

$\int \sec ^{2}(x)dx=\int du=u+C=\tan(x)+C.$

Example 2

Find $\int \tan(x)dx$ .

Let $u=\cos(x)$ so $du=-\sin(x)dx$ . Then

$\int \tan(x)dx$	$=\int {\frac {\sin(x)}{\cos(x)}}dx$
	$=\int -{\frac {du}{u}}$
	$=-\ln |u|+C$
	$=-\ln {\Big |}\cos(x){\Big |}+C$
	$=\ln {\Big |}\sec(x){\Big |}+C$

Example 3

Find $\int \sec(x)dx$ .

The trick to do this is to multiply and divide by the same thing like this:

$\int \sec(x)dx$	$=\int \sec(x){\frac {\sec(x)+\tan(x)}{\sec(x)+\tan(x)}}dx$
	$=\int {\frac {\sec ^{2}(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}}dx$

Making the substitution $u=\sec(x)+\tan(x)$ so $du=\sec(x)\tan(x)+\sec ^{2}(x)dx$ ,

$\int \sec(x)dx$	$=\int {\frac {du}{u}}$
	$=\ln |u|+C$
	$\ln {\Big |}\sec(x)+\tan(x){\Big |}+C$

More trigonometric combinations

For the integrals $\int \sin(nx)\cos(mx)dx$ or $\int \sin(nx)\sin(mx)dx$ or $\int \cos(nx)\cos(mx)dx$ use the identities

• $\sin(a)\cos(b)={\frac {\sin(a+b)+\sin(a-b)}{2}}$
• $\sin(a)\sin(b)={\frac {\cos(a-b)-\cos(a+b)}{2}}$
• $\cos(a)\cos(b)={\frac {\cos(a-b)+\cos(a+b)}{2}}$

Example 1

Find $\int \sin(3x)\cos(5x)dx$ .

We can use the fact that $\sin(a)\cos(b)={\frac {\sin(a+b)+\sin(a-b)}{2}}$ , so

$\sin(3x)\cos(5x)={\frac {\sin(8x)+\sin(-2x)}{2}}$

Now use the oddness property of $\sin(x)$ to simplify

$\sin(3x)\cos(5x)={\frac {\sin(8x)-\sin(2x)}{2}}$

And now we can integrate

$\int \sin(3x)\cos(5x)dx$	$=\int {\Big (}{\frac {\sin(8x)-\sin(2x)}{2}}{\Big )}dx$
	$={\frac {\cos(2x)}{4}}-{\frac {\cos(8x)}{16}}+C$

Example 2

Find:$\int \sin(x)\sin(2x)dx$ .

Using the identities

$\sin(x)\sin(2x)={\frac {\cos(-x)-\cos(3x)}{2}}={\frac {\cos(x)-\cos(3x)}{2}}$

Then

$\int \sin(x)\sin(2x)dx$	$={\frac {1}{2}}\int {\Big (}\cos(x)-\cos(3x){\Big )}dx$
	$={\frac {\sin(x)}{2}}-{\frac {\sin(3x)}{6}}+C$