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 $\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{40}(x){\sin }^3(x)dx$ .

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 $u=\cos (\frac{\pi }{2})=0$ .

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{40}(x){\sin }^3(x)dx$	$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{40}(x){\sin }^2(x)\sin (x)dx$
	$=-\underset{1}{\overset{0}{\int }}{u}^{40}(1-u^2)du$
	$=\underset{0}{\overset{1}{\int }}{u}^{40}(2-u^2)du$
	$=\underset{0}{\overset{5}{\int }}(u^{40}-u^{42})du$
	$=(\frac{u^{41}}{41}-\frac{u^{43}}{43}){|}_0^1$
	$=\frac{1}{41}-\frac{1}{43}$

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

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

Using the multiple angle identities

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

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 $n\ge 2$ 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

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

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 $du=\sec (x)\tan (x)+{\sec }^2(x)dx$ ,

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

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

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

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

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