## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 16

#### Answer

$$\frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \sqrt{\theta^{2}-1}+C$$

#### Work Step by Step

Given $$\int \theta \sec ^{-1} \theta d \theta$$ Let \begin{align*} u&= \sec ^{-1} \theta\ \ \ \ \ \ dv= \theta \, d\theta\\ du&= \frac{1}{|\theta| \sqrt{\theta^{2}-1}} d \theta\ \ \ \ \ \ v=\frac{1}{2}\theta^2 \end{align*} Then \begin{align*} \int x \csc x \cot x \, d x&= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \int \frac{\theta^2}{|\theta| \sqrt{\theta^{2}-1}} d \theta\\ &= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \int \frac{\theta }{ \sqrt{\theta^{2}-1}} d \theta\\ &= \frac{1}{2}\theta^2 \sec ^{-1} \theta-\frac{1}{2} \sqrt{\theta^{2}-1}+C \end{align*}

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