Integrals involving Powers of Trigonometric Functions

This wiki will enable you to evaluate integrals like $\displaystyle \int \tan^{4} x \, dx$ without splitting it up and doing lots of complicated integration.

For $n \geq 2$ and $n \in \mathbb{R}$,

$$\begin{aligned} \int \cos^{n} x \, dx &= \dfrac{1}{n} \cos^{n-1}x \sin x + \dfrac{n-1}{n}\int \cos^{n-2} x \, dx \\ \int \sin^{n} x \, dx &= -\dfrac{1}{n}\sin^{n-1}x \cos x + \dfrac{n-1}{n} \int \sin^{n-2}x \, dx. \end{aligned}$$

Proof: (Under construction)

Evaluate $\displaystyle \int \sin^{5} x \, dx$.

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For $n \geq 2$ and $n \in \mathbb{R}$,

$$\begin{aligned} \int \tan^{n} x \, dx &= \dfrac{\tan^{n-1}x}{n-1} - \int \tan^{n-2} x \, dx \\ \int \sec^{n} x \, dx &= \dfrac{\sec^{n-2} x \tan x}{n-1} + \dfrac{n-2}{n-1} \int \sec^{n-2} x \, dx. \end{aligned}$$

Proof: (Under construction)

Evaluate $\displaystyle \int \sec^{5} x \, dx$.

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There are a few possible scenarios for the integrand in the form $\sin^{\alpha} x \cos^{\beta} x$:

$$$$ Case 1: $\alpha$ is odd, $\beta$ is even

Evaluate $\displaystyle \int \sin^{3} x \cos^{2} x \, dx$.

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Rewrite it as $$\displaystyle \int \sin x \sin^{2} x \cos^{2} x \, dx = \int \sin x \big( 1-\cos^{2} x \big) \cos^{2} x \, dx. \qquad (1)$$ Let $u = \cos x,$ then $\frac{du}{dx} = -\sin x$ and $(1)$ can be rewritten as $$\displaystyle -\int \big( 1-u^{2} \big) u^{2} \, du.$$ This integral is very easy to compute now; you just have to expand the terms and use the power rule for antiderivatives. But don't forget what was $u$! $_\square$

$$$$ Case 2: $\alpha$ is even, $\beta$ is odd

Same idea as "$\alpha$ is odd, $\beta$ is even."

$$$$ Case 3: both $\alpha$ and $\beta$ even

In this scenario, there are two different things you could do. You could utilize the following identities:

• $\cos^{2} x = \frac{ 1+ \cos 2x}{2}$
• $\sin^{2} x = \frac{1 - \cos 2x}{2}.$

Or, you could rewrite the integrand only in terms of a single trigonometric function.

Evaluate $\displaystyle \int \sin^{2} x \cos^{2} x \, dx.$

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Using the aforementioned identities, we can rewrite the given expression as $$\begin{aligned} \int \left( \dfrac{1-\cos 2x}{2}\right) \left( \dfrac{1+ \cos 2x}{2} \right) \, dx &= \dfrac{1}{4} \displaystyle \int \big( 1- \cos^{2} 2x \big) \, dx \\ &= \dfrac{1}{4} \displaystyle \int \sin^{2} 2x \, dx. \qquad (1) \end{aligned}$$ Let $u = 2x,$ then$\frac{du}{dx} = 2$ and $(1)$ can be rewritten as $$\dfrac{1}{8} \displaystyle \int \sin^{2} u \, du. \qquad (2)$$ Remember? We got an identity for this! So $(2)$ can be rewritten as $$\dfrac{1}{8} \int \dfrac{1-\cos 2u}{2} \, du = \dfrac{u}{16} - \dfrac{\sin 2u}{32} + C,$$ where $C$ is the constant of integration.

But what was $u?$ Substituting $u=2x,$ the answer is $$\dfrac{x}{8} - \dfrac{\sin 4x}{32} + C. \ _\square$$

Something to consider: What's another way you could solve that integral? Since the powers are relatively nice, you could rewrite everything in terms of cosine or sine, and use one of the reduction formulas from above.

Evaluate $\displaystyle \int \sin^{2} x \cos^{4} x \, dx.$

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It is possible to use the aforementioned problems to solve this problem, but I'm going to use the reduction formulas approach just to contrast the methods. For this type of problem, I actually prefer the reduction formulas: $$\begin{aligned} \int \sin^{2} x \cos^{4} x \, dx & = \int \big( 1- \cos^{2} x \big) \cos^{4} x \, dx \\ & = \int \cos^{4} x \, dx - \int \cos^{6} x \, dx. \qquad (1) \end{aligned}$$ Since $\displaystyle \int \cos^{6} \, dx,$ which is the integral with the highest exponent in $(1),$ can be reduced to $$\int \cos^{6} \, dx=\dfrac{\cos^{5} x \sin x}{6} + \dfrac{5}{6} \displaystyle \int \cos^{4} x \, dx, \qquad (2)$$ substituting $(2)$ into $(1)$ gives $$\int \sin^{2} x \cos^{4} x \, dx = \dfrac{1}{6} \displaystyle \int \cos^{4} \, dx - \dfrac{\cos^{5} x \sin x}{6}. \qquad (3)$$

Use the reduction formulas again, then $\displaystyle \int \cos^{4} x \, dx$ in $(3)$ becomes $$\int \cos^{4} \, dx = \dfrac{\cos^{3} x \sin x }{4} + \dfrac{3}{4} \int \cos^{2} x \, dx. \qquad (4)$$ We know that $\cos^{2} x = \frac{1+ \cos 2x}{2}$ and we also know how to integrate that, so I'm going to skip that step and find $(4)$ to be equivalent to $$\int \cos^{4} \, dx = \dfrac{\cos^{3} x \sin x }{4} + \dfrac{3}{4} \left( \dfrac{x}{2} + \dfrac{\sin 2x}{4} \right). \qquad (5)$$ Substituting $(5)$ into $(3)$ gives $$\int \sin^{2} x \cos^{4} x \, dx =\dfrac{\cos^{3} x \sin x}{24} + \dfrac{x}{16} + \dfrac{\sin 2x}{32} - \dfrac{\cos^{5} x \sin x}{6} + C,$$ where $C$ is the constant of integration. $_\square$