## Solving integrals of the form sinm x cosn x when m and n are even

In my last post, I showed you the method for integrating functions of the form

$\int \sin^m{x} \; \cos^n{x} \; dx$

when either m or n is odd. If they are both even, the process is more complicated. In this post, I’ll show you how to attack this problem. There are two different methods you can use; if one doesn’t work so well, try the other. We will demonstrate both methods here with an example. Let’s integrate

$\int \sin^2{x} \; \cos^4{x} \; dx$

Method 1

First, rewrite the integral so that the factors are powers of sinx and cosx.

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

Use the power reducing formulas

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

to replace both the sinx and the cosx terms.

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

Then expand each term and multiply it all out (hey, I said it was more complicated…)

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

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

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

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

Now we are finally ready to integrate. The first two terms can be integrated by inspection. The third term is attacked by using the power reducing formula. In the last term, the exponent is odd, so apply the technique we used in the last blog post:

$\displaystyle \frac{1}{8} \int 1 + \cos{2x} - \frac{1}{2} (1+\cos{4x}) - \cos{2x}\cos^2{2x} \; dx =$

$\displaystyle \frac{1}{8} \int 1 + \cos{2x} - \frac{1}{2} (1+\cos{4x}) - \cos{2x}(1- \sin^2{2x}) \; dx =$

$\displaystyle \frac{1}{8} \int \frac{1}{2} - \frac{1}{2} \cos{4x} + \cos{2x} \sin^2{2x} \; dx =$

$\displaystyle \frac{1}{8} \left( \frac{1}{2}x - \frac{1}{8} \sin{4x} + \frac{1}{6} \sin^3{2x} \right) + C = \frac{1}{16}x - \frac{1}{64} \sin{4x} + \frac{1}{48} \sin^3{2x} + C$

Whew!

Method 2:

First, rearrange terms so that there is a power of (sin x cos x).

$\displaystyle \int \sin^2{x} \cos^4{x} \; dx = \int \sin^2{x} \cos^2{x} \cos^2{x} \; dx =$

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

Then use the double angle formula $\sin{x} \cos{x} = \frac{1}{2} \sin{2x}$ and the power reducing formulas as necessary:

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

Expand the terms and multiply out:

$\displaystyle \int \left(\frac{1}{2} \sin{2x} \right)^2 \left(\frac{1}{2} (1+ \cos{2x}) \right) \; dx = \frac{1}{8} \int \sin^2{2x} (1+ \cos{2x}) \; dx =$

$\displaystyle \frac{1}{8} \int \sin^2{2x} + \sin^2{2x} \cos{2x} \; dx = \frac{1}{8} \int \frac{1}{2} (1- \cos{4x}) + \sin^2{2x} \cos{2x} \; dx =$

$\displaystyle \frac{1}{8} \left( \frac{1}{2}x - \frac{1}{8} \sin{4x} + \frac{1}{6} \sin^3{2x} \right) + C = \frac{1}{16}x - \frac{1}{64} \sin{4x} + \frac{1}{48} \sin^3{2x} + C$

Note that this is the same result as in Method 1 above.

Calculus
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