Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 112


$$\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan (x))+c$$

Work Step by Step

Given $$\int \frac{d x}{1+\sin ^{2} x}$$ Let $u=\tan x\ \ \ \ \ \ du=\sec^2 xdx $, then \begin{align*} dx&=\frac{du}{\sec^2 x}\\ &= \frac{du}{1+\tan^2 x}\\ &=\frac{du}{1+u^2} \end{align*} Since \begin{align*} \sin ^{2}(x)&=\frac{\tan ^{2} x}{1+\tan ^{2} x}\\ &=\frac{u^{2}}{1+2u^{2}} \end{align*} Then \begin{align*} \int \frac{d x}{1+\sin ^{2} x}&= \int \frac{1+u^{2}}{1+2 u^{2}} \cdot \frac{d u}{1+u^{2}}\\ &=\int \frac{d u}{1+2 u^{2}}\\ &=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} u)+c\\ &=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan (x))+c \end{align*}
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