Evaluate the integral _ 0 ^ 1 ^ -1 (2 x 1+x^ 2 ) d x using substitution.

Given: _ 0 ^ 1 ^ -1 (2 x x^ 2 +1 ) d x Let x = tan dx = ^2 d When, x = 0, = 0 and when x = 1, = 4 Let I= _ 0 ^ 1 ^ -1 (2 x x^ 2 +1 ) d x I= _ 0 ^ 4 ^ -1 (2 ^ 2 +1 ) ^ 2 d = _ 0 ^ 4 ^ -1 ( 2 ) ^ 2 d = _ 0 ^ 4 2 ^ 2 d = 2 _ 0 ^ 4 ^ 2 d By applying product rule as, u v d x=u vdx- du dx vdx d x I=2 [ ^ 2 d - d d ^ 2 d d ]_ 0 ^ 4 = 2 [ - 1 . d ]_ 0 ^ 4 = 2[ - | |]_ 0 ^ 4 = 2 [ 4 4 - | 4 |-0+ | | ] = 2 [ 4 -1 2 (2) .] = 2 - (2)

Evaluate the integral _ 0 ^ 1 ^ -1 (2 x 1+x^ 2 ) d x using substi…

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Question

Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$ using substitution.

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Answer

Given: $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$
Let x = tan $\theta$ $\Rightarrow dx = \sec^2 \theta$ d$\theta$
When, x = 0, $\theta$ = 0 and when x = 1, $\theta = \frac{\pi}{4}$
Let $I=\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{x^{2}+1}\right) d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \sin ^{-1}\left(\frac{2 \tan \theta}{\tan ^{2} \theta+1}\right) \sec ^{2} \theta \mathrm{d} \theta$
= $\int_{0}^{\frac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta$
= $\int_{0}^{\frac{\pi}{4}} 2 \theta \sec ^{2} \theta d \theta$
= $2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta$
By applying product rule as,
$\int \mathrm{u} \cdot \mathrm{v} \mathrm{d} \mathrm{x}=\mathrm{u} \cdot \int \mathrm{vdx}-\int \frac{\mathrm{du}}{\mathrm{dx}} \cdot\left\{\int \mathrm{vdx}\right\} \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=2\left[\theta \int \sec ^{2} \theta \mathrm{d} \theta-\int \frac{\mathrm{d}}{\mathrm{d} \theta} \theta \cdot\left\{\int \sec ^{2} \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]_{0}^{\frac{\pi}{4}}$
= $2\left[\theta \tan \theta-\int 1 . \tan \theta d \theta\right]_{0}^{\frac{\pi}{4}}$
= $2[\theta \tan \theta-\log |\sec \theta|]_{0}^{\frac{\pi}{4}}$
= $2\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\log \left|\sec \frac{\pi}{4}\right|-0+\log |\sec \theta|\right]$
= $2\left[\frac{\pi}{4}-\frac{1}{2} \log (2)\right.]$
= $\frac{\pi}{2}-\log (2)$

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