## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 44

#### Answer

$$\frac{-1}{x}\sin^{-1}x-\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c$$

#### Work Step by Step

Given $$\int \frac{\sin^{-1}x}{x^2}dx$$ Use integration by parts , let \begin{aligned} u&= \sin^{-1}x\ \ \ \ \ \ \ &dv&= x^{-2}dx\\ du&= \frac{dx}{\sqrt{1-x^2}}\ \ \ \ \ \ \ & v&= \frac{-1}{x} \end{aligned} Then \begin{aligned} \int \frac{\sin^{-1}x}{x^2}dx&= \frac{-1}{x}\sin^{-1}x+\int \frac{dx}{x\sqrt{1-x^2}} \end{aligned} To evaluate $\int \dfrac{dx}{x\sqrt{1-x^2}}$, let $$x=\sin \theta \ \ \ \ \ dx=\cos \theta d\theta$$ Then \begin{aligned} \int \dfrac{dx}{x\sqrt{1-x^2}}&=\int \frac{\cos \theta d\theta }{\sin \theta \sqrt{1-\sin^2\theta }}\\ &=\int \frac{1}{\sin \theta } d\theta \\ &=\int \csc \theta d\theta \\ &=-\ln |\csc \theta +\cot \theta |+c\\ &= -\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c \end{aligned} Hence \begin{aligned} \int \frac{\sin^{-1}x}{x^2}dx&= \frac{-1}{x}\sin^{-1}x-\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c \end{aligned}

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