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Expert-verified Found in: Page 326 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Evaluate the Integral: $$\int {{{\cot }^5}\theta {{\sin }^4}} \theta d\theta$$

To evaluate this integral, we will use trigonometric identities $$\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$ and $${\sin ^2}\theta + {\cos ^2}\theta = 1$$

Firstly, we will use $$\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$ to change the integral in sine and cosine functions. Then we will use $${\sin ^2}\theta + {\cos ^2}\theta = 1$$ followed by substitution method

$$= \int {{{\cot }^5}} \theta {\sin ^4}\theta d\theta = \ln |\sin \theta | - {\sin ^2}\theta + \frac{1}{4}{\sin ^4}\theta + c$$

See the step by step solution

## Step 1: Using Trigonometric Identities

Let $$I = \int {{{\cot }^5}\theta {{\sin }^4}} \theta d\theta$$

As $$\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$

\begin{aligned}{l}I &= \int {\frac{{{{\cos }^5}\theta }}{{{{\sin }^5}\theta }}} {\sin ^4}\theta d\theta \\ &= \int {\frac{{{{\cos }^5}\theta }}{{\sin \theta }}} d\theta \end{aligned}

## Step 2: Splitting of fifth power of cosine function

Split $${\cos ^5}\theta$$ as $$\cos \theta {\cos ^4}\theta$$ and use $${\cos ^2}\theta = 1 - {\sin ^2}\theta$$

\begin{aligned}{l}I &= \int {\frac{{{{\cos }^4}\theta \cos \theta }}{{\sin \theta }}} d\theta \\ &= \int {\frac{{{{(1 - {{\sin }^2}\theta )}^2}}}{{\sin \theta }}} \cos \theta d\theta \end{aligned}

## Step 3: Substitution Method

Let \begin{aligned}{l}\sin \theta &= t\\ &= \cos \theta d\theta &= dt\\I &= \int {\frac{{{{(1 - {t^2})}^2}}}{t}} dt\end{aligned}

## Step 4: Using Square identity formula

As $${(a - b)^2} = {a^2} + {b^2} - 2ab$$

\begin{aligned}{l}I &= \int {\frac{{{t^4} - 2{t^2}}}{t}} dt\\ &= \int {(\frac{1}{t}} - 2t + {t^3})dt\end{aligned}

## Step 5: Splitting Of Integral

Split the integral and use $$\int {\frac{1}{\theta }} d\theta = \ln |\theta |$$

\begin{aligned}{l}I &= \int {\frac{1}{t}} dt - 2\int {tdt + \int {{t^3}} } dt\\ &= \ln |t| - {t^2} + \frac{{{t^4}}}{4} + c\end{aligned}

## Step 6: Re-Substitution of the value of t

None Re-Substitute to get the answer

$$I = ln|\sin \theta | - {\sin ^2}\theta + \frac{{{{\sin }^4}\theta }}{4} + c$$

Hence, the value of integral

$$\int {{{\cot }^5}} \theta {\sin ^4}\theta d\theta = \ln |\sin \theta | - {\sin ^2}\theta + \frac{{{{\sin }^4}\theta }}{4} + c$$ ### Want to see more solutions like these? 