Advanced Trigonometric Function Integration Examples 2

# Advanced Trigonometric Function Integration 2

We will continue onward to integrating advanced trigonometric functions. We have already looked at three strategies on the Advanced Trigonometric Function Integration Examples 1 page.

## Example 1

Evaluate the following integral: $\int \sin ^4 x \cos ^2 x \: dx$

Unlike previous examples, neither of the exponents are even. Hence, we will need to rewrite this integral in terms of one trigonometric function only and see what we can do from there. Let's rewrite the integral in terms of cosx only:

(1)
\begin{align} \int \sin ^4 x \cos ^4 x \: dx = \int (\sin ^2 x)^2 \cos ^2 x \: dx \\ \int \sin ^4 x \cos ^4 x \: dx = \int (1 - \cos ^2 x)^2 \cos ^2 x \: dx \\ \int \sin ^4 x \cos ^4 x \: dx = \int (1 - 2\cos ^2 x + \cos ^4 x) \cos ^2 x \: dx \\ \int \sin ^4 x \cos ^4 x \: dx = \int \cos ^2 x - 2\cos ^4 x + \cos ^6 x \: dx \\ \quad \int \sin ^4 x \cos ^4 x \: dx = \int \cos ^2 x \: dx -2 \int \cos ^4 x \: dx + \int \cos ^6 x \: dx \\ \end{align}

We can now evaluate this integral with rather tedious methods we've learned before. In fact, we're going to use the following reduction formula:

(2)
\begin{align} \quad \int cos^n x \: dx = \frac{1}{n} \cos ^{n - 1} x \sin x + \frac{n - 1}{n} \int \cos ^{n-2} x \: dx \end{align}

We thus obtain:

(3)
\begin{align} \quad \quad \int \sin ^4 x \cos ^4x \: dx = \frac{1}{2} \cos x \sin x + \frac{x}{2} - 2 [\frac{1}{4} \cos ^{3} x \sin x + \frac{3}{4} \int \cos ^{2} x \: dx] + \frac{1}{6} \cos ^{5} x \sin x + \frac{5}{6} \int \cos ^{4} x \: dx \\ \quad \quad \int \sin ^4 x \cos ^4x \: dx = \frac{1}{2} \cos x \sin x + \frac{x}{2} - \frac{1}{2} \cos ^{3} x \sin x - \frac{3}{2} \int \cos ^{2} x \: dx + \frac{1}{6} \cos ^{5} x \sin x + \frac{5}{6} \int \cos ^{4} x \: dx \\ \quad \quad \quad \int \sin ^4 x \cos ^4x \: dx = \frac{1}{2} \cos x \sin x + \frac{x}{2} - \frac{1}{2} \cos ^{3} x \sin x - \frac{3}{2} [ \frac{1}{2} \cos x \sin x + \frac{x}{2} ] + \frac{1}{6} \cos ^{5} x \sin x + \frac{5}{6} [ \frac{1}{4} \cos ^{3} x \sin x + \frac{3}{4} \int \cos ^{2} x \: dx] \\ \quad\quad \quad \int \sin ^4 x \cos ^4x \: dx = \frac{1}{2} \cos x \sin x + \frac{x}{2} - \frac{1}{2} \cos ^{3} x \sin x - \frac{3}{2} [ \frac{1}{2} \cos x \sin x + \frac{x}{2} ] + \frac{1}{6} \cos ^{5} x \sin x + \frac{5}{6} [ \frac{1}{4} \cos ^{3} x \sin x + \frac{3}{4} [\frac{1}{2} \cos x \sin x + \frac{x}{2}]] \\ \end{align}

# Trigonometric Products Containing Even Powers of Sine or Cosine.

As seen in Example 1, using reduction formulas is sometimes very helpful for evaluating these integrals, however, there are other methods. Let's look at the following strategy:

 Strategy 4: If $f(x) = \sin ^m x$, $f(x) = \cos ^n x$ or $f(x) = \sin ^m x \cos ^m x$ where both $m$ and $n$ are even, then use the half-angle trigonometric identities $\sin ^2 x = \frac{1}{2}(1 - \cos 2x)$, $\cos ^2 x = \frac{1}{2} (1 + \cos 2x )$, and $\sin x \cos x = \frac{1}{2} \sin 2x$ where necessary.

## Example 2

Evaluate the following integral: $\int \sin ^2 x \cos ^2 x \: dx$.

By trigonometric substitution we obtain that:

(4)
\begin{align} \quad \int \sin ^2 x \cos ^2 x \: dx = \int \frac{1}{2}(1 - \cos 2x) \cdot \frac{1}{2} (1 + \cos 2x) \: dx \\ \int \sin ^2 x \cos ^2 x \: dx = \int \frac{1}{4}(1 - \cos 2x)(1 + \cos 2x) \: dx \\ \int \sin ^2 x \cos ^2 x \: dx = \int \frac{1}{4}(1 - \cos ^2 2x) \: dx \\ \int \sin ^2 x \cos ^2 x \: dx = \frac{x}{4} - \frac{1}{4} \int \cos ^2 2x \: dx \\ \int \sin ^2 x \cos ^2 x \: dx = \frac{x}{4} - \frac{1}{4} \int \frac{1}{2}(1 + \cos 4x) \: dx \\ \int \sin ^2 x \cos ^2 x \: dx = \frac{x}{4} - \frac{x}{8} - \frac{\sin 4x}{32} + C \\ \int \sin ^2 x \cos ^2 x \: dx = \frac{x}{8} - \frac{\sin 4x}{32} + C \end{align}

# Trigonometric Products of Tan and Secant where the Power of Secant is Even

 Strategy 5: If $f(x) = \sec ^n x$ or $f(x) = \tan ^m x \sec ^n x$ where $n$ is even, then factor out a $\sec ^2 x$. Use the trigonometric identity $\sec ^2 x = 1 + \tan ^2 x$ and let $u = \tan x$.

## Example 3

Evaluate the following integral: $\int \tan ^6 x \sec ^4 x \: dx$.

First let's factor out a $\sec ^2 x$ to obtain:

(5)
\begin{align} \int \tan ^6 x \sec ^4 x \: dx = \int \tan ^6 x \sec ^2 x \sec ^2 x \: dx \end{align}

Then make the trigonometric substitution prescribed above and integrate with substitution by letting $u = \tan x$ and $du = \sec ^2 x$.

(6)
\begin{align} \int \tan ^6 x \sec ^4 x \: dx = \int \tan ^6 x (1 + \tan ^2 x) \sec ^2 x \: dx \\ \int \tan ^6 x \sec ^4 x \: dx = \int u^6 (1 + u^2) \: du \\ \int \tan ^6 x \sec ^4 x \: dx = \int u^6 + u^8 \: du \\ \int \tan ^6 x \sec ^4 x \: dx = \frac{u^7}{7} + \frac{u^9}{9} + C\\ \int \tan ^6 x \sec ^4 x \: dx = \frac{\tan ^7 x}{7} + \frac{\tan ^9 x}{9} + C \\ \end{align}

# Trigonometric Products of Tan and Secant where the Power of Tan is Odd

 Strategy 6: If $f(x) = \tan ^m x$ or $f(x) = \tan ^m x \sec ^n x$ where $m$ is odd, then factor out a $\sec x \tan x$. Use the trigonometric identity $\tan ^2 x = \sec ^2 x - 1$ and let $u = \sec x$.

## Example 4

Evaluate the following integral: $\int \tan ^5 x \sec ^ 7 x \: dx$.

We will follow Strategy 6 exactly to evaluate this integral:

(7)
\begin{align} \quad \int \tan ^5 x \sec ^ 7 x \: dx = \int \tan ^4 x \sec ^ 6 x (\tan x \sec x) \: dx \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \int (\sec ^2 x - 1)^2 \sec ^ 6 x (\tan x \sec x) \: dx \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \int (u^2 - 1)^2 u^6 \: du \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \int (u^4 - 2u^2 + 1) u^6 \: du \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \int u^{10} - 2u^8 + u^6 \: du \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \frac{u^{11}}{11} - \frac{2u^9}{9} + \frac{u^7}{7} + C \\ \quad \int \tan ^5 x \sec ^ 7 x \: dx = \frac{\sec ^{11} x}{11} - \frac{2\sec ^9 x}{9} + \frac{\sec ^7 x}{7} + C \\ \end{align}