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 $\cos^m(x)\sin^n(x)$. 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

$\begin{matrix} \int\cos^3(x)\sin^2(x)\,dx &=&\int\cos^2(x)\sin^2(x)\cos(x)\,dx\\ &=&\int (1-u^2)u^2\,du\\ &=&\int u^2\,du - \int u^4\,du\\ &=&{1\over 3} u^3+{1\over 5}u^5 + C\\ &=&{1\over 3} \sin^3(x)-{1\over 5}\sin^5(x)+C \end{matrix}.$

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 $\int_0^{\pi/2} \cos^{40}(x)\sin^3(x) dx$.

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 $x=\pi/2$ we have $u=\cos(\pi/2) = 0$.

$\begin{matrix} \int_0^{\pi/2} \cos^{40}(x)\sin^3(x) dx &=& \int_0^{\pi/2} \cos^{40}(x)\sin^2(x) \sin(x) dx \\ &=& -\int_{1}^{0} u^{40} (1-u^2) du \\ &=&\int_{0}^{1} u^{40} (1-u^2) du\\ &=& \int_{0}^{1} u^{40} - u^{42} du \\ &=& [\frac{1}{41}u^{41} - \frac{1}{43}u^{43}]_0^1 \\ &=& \frac{1}{41}-\frac{1}{43}. \end{matrix}$

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

Example

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

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

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

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

$\begin{matrix} I & = & \frac{1}{8} \left( \int 1 \, dx - \int \cos^2(2x)\, dx + \int \cos(2x)\,dx -\int \cos^3(2x)\,dx \right) \\ & = & \frac{1}{8} \left( x - \frac{1}{2} \int (1 + \cos(4x))\,dx + \frac{1}{2}\sin(2x) -\int \cos^2(2x) \cos(2x) \,dx\right) \\ & = & \frac{1}{16} \left( x + \sin(2x) + \int \cos(4x) \,dx -2 \int(1-\sin^2(2x))\cos(2x)\,dx\right) \\ \end{matrix}$

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

$\begin{matrix} \int \tan(x)dx &=& \int \frac{\sin(x)}{\cos(x)} dx \\ &=& \int \frac{-1}{u} du \\ &=& -\ln |u| + C \\ &=& -\ln |\cos(x) | + C\\ &=& \ln |\sec(x)| +C. \end{matrix}$

Example 3

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

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

$\begin{matrix} \int \sec(x)dx &=& \int \sec(x)\frac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} dx \\ &=& \int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x)+ \tan(x)} \end{matrix}.$

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

$\begin{matrix} \int \sec(x) dx &=& \int \frac{1}{u} du\\ &=& \ln |u| + C \\ &=& \ln |\sec(x) + \tan(x)| + C \end{matrix}.$

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

• $\sin(a)\cos(b) = {1\over 2}(\sin{(a+b)}+\sin{(a-b)}) \,$
• $\sin(a)\sin(b) = {1\over 2}(\cos{(a-b)}-\cos{(a+b)}) \,$
• $\cos(a)\cos(b) = {1\over 2}(\cos{(a-b)}+\cos{(a+b)}) \,$

Example 1

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

We can use the fact that $\sin(a)\cos(b)=(1/2)(\sin(a+b)+\sin(a-b))$, so

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

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

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

And now we can integrate

$\begin{matrix} \int \sin(3x)\cos(5x)\,dx & = & \frac{1}{2} \int \sin(8x)-\sin(2x)dx \\ & = & \frac{1}{2}(-\frac{1}{8}\cos(8x)+\frac{1}{2}\cos(2x)) +C \\ \end{matrix}$

Example 2

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

Using the identities

$\sin(x) \sin(2x)= \frac{1}{2} \left( \cos(-x)-\cos(3x) \right) = \frac{1}{2} (\cos(x) -\cos(3x)).$

Then

$\begin{matrix} \int \sin(x)\sin(2x)\,dx & = & \frac{1}{2} \int (\cos(x)-\cos(3x))\,dx \\ & = & \frac{1}{2}(\sin(x)-\frac{1}{3}\sin(3x)) + C \end{matrix}$