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Integrals Involving Powers of Sine and Cosine

Integrals involving powers, such as $\int\mathrm{\sin^m{(x)}\cos^n{(x)}}\, \mathrm{d}x$

For example, if you want to evaluate $\int\mathrm{\sin^5{(x)}\cos{(x)}}\, \mathrm{d}x$, you let $u = \sin{(x)}$, so the integral becomes $\int\mathrm{u^5}\, \mathrm{d}u = \frac{u^6}{6} + C = \frac{\sin^6{(x)}}{6} + C$


OIGOIG means "Odd is good. Odd is great." So, when one of your powers is odd and the other is even, that's really easy to do. 

Odd Sine Power

For example, $$\int\mathrm{\sin^{2k + 1}{(x)}\cos^n{(x)}}\,\mathrm{d}x$$ $$\int\mathrm{(sin^2{(x)})^k\cos^n{(x)}sin{(x)}}\, \mathrm{d}x$$ $$\int\mathrm{(1-\cos^2{(x)})^k\cos^n{(x)}\sin{(x)}}\, \mathrm{d}x$$ Note that $\sin{(x)}dx$ is the du, so $u = \cos{(x)}$ when $\sin$ is the odd power. $u = \cos{(x)}$, $du = -\sin{(x)}dx$ So, it becomes $$-\int\mathrm{(1- u^2)^ku^n}\,\mathrm{d}u%$$

Odd Cosine Power

$$\int\mathrm{\cos^{2k + 1}{(x)}\sin^n{(x)}}\, \mathrm{d}x = \int\mathrm{(\cos^2{(x)})^k\sin^n{(x)}\cos{(x)}}\, \mathrm{d}x$$ So $\cos{(x)}dx = du$, so $u = \sin{(x)}$ $$\int\mathrm{(1 - u^2)^ku^n}\, \mathrm{d}u$$

So, for both of those cases, if one of the powers is odd, you take out one of the functions and put it at the end, so in odd cosine, you'd put the cosine at the end. In odd sine, you'd put the sine at the end.

Both Even

That sucks. You have to power reduce, meaning you turn $\sin^2{x}$ into $\frac{1 - \cos{(2x)}}{2}$ and $\cos^2{(x)}$ into $\frac{1 + \cos{2x}}{2}$.


Example 1

$$\int\mathrm{\sin^3{x}\cos^4{x}}\, \mathrm{d}x = \int\mathrm{\sin^2{x}\cos^4{x}\sin{x}}\, \mathrm{d}x$$ $$u = \cos{(x)}, du = -\sin{(x)}$$ $$=-\int\mathrm{(1 - u^2)u^4}\, \mathrm{d}u$$ $$=-\int\mathrm{u^4 - u^6}\, \mathrm{d}u$$ $$= -\frac{u^5}{5} + \frac{u^7}{7} + C$$ $$=-\frac{\cos^5{(x)}}{5} + \frac{\cos^7{(x)}}{7} + C$$

Example 2

$$\int\mathrm{\sin^4{(x)}\cos^3{(x)}}\, \mathrm{d}x$$ $$\int\mathrm{\sin^4{(x)}\cos^2{(x)}\cos{(x)}}\, \mathrm{d}x$$ $$du = \cos{(x)}dx, u = \sin{(x)}$$ $$\int\mathrm{u^4(1-u^2)}\, \mathrm{d}u$$ $$\int\mathrm{u^4 - u^6}\, \mathrm{d}u$$ $$\frac{u^5}{5} - \frac{u^7}{7} + C$$ $$\frac{\sin^5{x}}{5} - \frac{\sin^7{x}}{7} + C$$


This stands for "Except for u subs of tangents"

Even Secant

$$\int\mathrm{\sec^{2k}{(x)}\tan^n{(x)}}\, \mathrm{d}x$$ $$\int\mathrm{\sec^{2k-2}{(x)}\tan^n{(x)}\sec^2{(x)}}\, \mathrm{d}x$$ $$du = \sec^2{(x)}dx, u = \tan{(x)}$$ $$\int\mathrm{(\sec^2{(x)})^{k-1}u^n}\, \mathrm{d}u$$ $$\int\mathrm{(1 + u^2)^{k-1}u^n}\, \mathrm{d}u$$

Odd Tangent

$$\int\mathrm{\sec^m{(x)}\tan^{2k + 1}{(x)}}\, \mathrm{d}x$$ $$\int\mathrm{\sec^{m-1}{(x)}\tan^{2k}{(x)}\sec{(x)}\tan{(x)}}\, \mathrm{d}x$$ $$du = \sec{(x)}\tan{(x)}dx, u = \sec{(x)}$$ $$\int\mathrm{u^{m-1}(u^2 - 1)^k}\, \mathrm{d}x$$

Only Tangent

This is kinda recursive.

$$\int\mathrm{\tan^n{(x)}}\, \mathrm{d}x = \int\mathrm{\tan^{n-2}{(x)}(sec^2{(x)} - 1)}\, \mathrm{d}x$$ $$\int\mathrm{\tan^{n-2}{(x)}\sec^2{(x)}}\,\mathrm{d}x - \int\mathrm{\tan^{n-2}{(x)}}\, \mathrm{d}x$$ $$=\frac{tan^{n-1}{(x)}}{n-1} - \int\mathrm{\tan^{n-2}{(x)}}\, \mathrm{d}x$$

(note that the first part was done using a u-sub)
The recursive part of the solution is you have to keep doing that until you reduce the power of the tan to ∫tan2(x)dx (just convert to sec) or ∫tan(x)dx (u-sub).

David Witten

Integration by Parts

(Not) The Second Derivative Test