Advanced Trigonometric Function Integration Examples 1

# Advanced Trigonometric Function Integration Examples 1

We're going to continue looking through examples of Advanced Trigonometric Function Integration, and look at some strategies to solving them. We will look at three more strategies on the Advanced Trigonometric Function Integration Examples 2 page.

# Trigonometric Products Containing an Odd Power of Cosine

 Strategy 1: If $f(x) = \cos ^n x$ or $f(x) = \sin ^m x \cos ^n x$ and $n$ is odd, then factor out a $\cos x$. Use the trigonometric identity $\cos ^2 x = 1 - \sin ^2 x$, and then use the substitution $u = \sin x$.

## Example 1

Evaluate the following integral $\int \cos ^5 x \: dx$.

Let's first separate one of the cosines and use the identity $\cos ^2 x = 1 - \sin ^2 x$ to obtain:

(1)
\begin{align} \int \cos ^5 x \: dx = \int \cos ^4 x \cos x \: dx \\ \int \cos ^5 x \: dx = \int (\cos ^2 x)^2 \cos x \: dx \\ \int \cos ^5 x \: dx = \int (1 - \sin ^2 x)^2 \cos x \: dx \end{align}

Now let's integrate this by substitution. Let's let $u = \sin x$. Then $du = \cos x \: dx$. Now let's apply our substitution:

(2)
\begin{align} \int \cos ^5 x \: dx = \int (1 - u^2)^2 \: du \\ \int \cos ^5 x \: dx = \int 1 - 2u^2 + u^4 \: du \\ \int \cos ^5 x \: dx = u - \frac{2u^3}{3} + \frac{u^5}{5} + C \\ \int \cos ^5 x \: dx = \sin x - \frac{2\sin ^3 x}{3} + \frac{\sin ^5 x}{5} + C \\ \end{align}

The technique is rather simply, and can be applied to integrating all functions in the form $f(x) = \cos ^ m x \: dx$ where m is an odd integer ≥ 3.

## Example 2

Evaluate the following integral: $\int \cos ^7 x \: dx$.

Once again we will work in the same procedure as Example 1 with the same substitution of $u = \sin x$ and $du = \cos x \: dx$.

(3)
\begin{align} \int \cos ^7 x \: dx = \int \cos ^6 x \cos x \: dx \\ \int \cos ^7 x \: dx = \int (\cos ^2 x)^3 \cos x \: dx \\ \int \cos ^7 x \: dx = \int (1 - \sin ^2 x)^3 \cos x \: dx \\ \int \cos ^7 x \: dx = \int (1 - u)^3 \: du \\ \int \cos ^7 x \: dx = \int (1 - u)^3 \: du \\ \quad \int \cos ^7 x \: dx = \int 1 - u - 2u^2 + 2u^3 + u^4 - u^5 \: du \\ \quad \int \cos ^7 x \: dx = u - \frac{u^2}{2} - \frac{2u^3}{3} + \frac{2u^4}{4} + \frac{u^5}{5} - \frac{u^6}{6} + C \\ \quad \int \cos ^7 x \: dx = \sin x - \frac{\sin^2 x}{2} - \frac{2\sin ^3 x}{3} + \frac{\sin ^4 x}{2} + \frac{\sin ^5 x}{5} - \frac{\sin ^6 x}{6} + C \\ \end{align}

# Trigonometric Products Containing an Odd Power of Sine

 Strategy 2: If $f(x) = \sin ^m x$ or $f(x) = \sin ^m x \cos ^n x$ and $m$ is odd, then factor out a $\sin x$. Use the trigonometric identity $\sin ^2 x = 1 - \cos ^2 x$, and then use the substitution $u = \cos x$.

## Example 3

Evaluate the following integral $\int \sin ^5 x \: dx$.

This time let's separate one sine out and make the substitution $u = \cos x$ and $du = -\sin x \: dx$. The idea is identical to the last two examples.

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

## Example 4

Evaluate the following integral: $\int \cos ^9 x \sin ^3 x \: dx$.

Since the power of cosine and sine are both odd, we can apply strategy 1 or strategy 2. Let's apply strategy 2. Once again let's reconfigure the function and use the the substitution that $u = \cos x$ and $du = -\sin x \: dx$.

(5)
\begin{align} \int \cos ^9 x \sin ^3 \: dx = \int \cos ^9 x \sin ^2 x \sin x \: dx \\ \int \cos ^9 x \sin ^3 \: dx = \int \cos ^9 x (1 - \cos ^2 x) \sin x \: dx \\ \int \cos ^9 x \sin ^3 \: dx = - \int u^9 (1 - u^2) \: du \\ \int \cos ^9 x \sin ^3 \: dx = - \int u^9 - u^11 \: du \\ \int \cos ^9 x \sin ^3 \: dx = - (\frac{u^{10}}{10} - \frac{u^{12}}{12}) + C \\ \int \cos ^9 x \sin ^3 \: dx = - \left (\frac{u^{10}}{10} + \frac{u^{12}}{12} \right ) + C\\ \int \cos ^9 x \sin ^3 \: dx = - \frac{\cos ^{10} x}{10} + \frac{\cos ^{12} x}{12} + C\\ \end{align}

# Trigonometric Products of Sin(ax)Cos(bx)

 Strategy 3: If $f(x) = \sin (ax) \sin (bx)$, $f(x) = \sin (ax) \cos (bx)$ or $f(x) = \cos (ax) \cos (bx)$, then $\int f(x) \: dx$ can be evaluated using one of the following trigonometric identities: 1. $\sin (ax) \sin (bx) = \frac{1}{2} [ \cos (ax - bx) - \cos (ax + bx) ]$ 2. $\sin (ax) \cos (bx) = \frac{1}{2} [ \sin (ax - bx) + \sin (ax + bx) ]$ 3. $\cos (ax) \cos (bx) = \frac{1}{2} [ \cos (ax - bx) + \cos (ax + bx)$.

## Example 5

Evaluate the following integral: $\int \sin 4x \cos 3x \: dx$.

Using the second identity above, we get that:

(6)
\begin{align} \int \sin 4x \cos 3x \: dx = \frac{1}{2} \int \sin x + \sin 7x \: dx \\ \: \int \sin 4x \cos 3x \: dx = \frac{1}{2} \left ( -\cos x -\frac{1}{7}\cos 7x \right ) + C \end{align}