_0^1 , ^ - 1 ,x x ,dx = - Vidyadip

MCQ

$\int\limits_0^1 {\,\frac{{{{\tan }^{ - 1}}\,x}}{x}\,dx} $ $=$

A
$\int\limits_0^{\frac{\pi }{4}} {\,\frac{x}{{\sin x}}\,dx} $
B
$\int\limits_0^{\frac{\pi }{2}} {\,\frac{x}{{\sin x}}\,dx} $
✓
$\frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\,\frac{x}{{\sin x}}\,dx} $
D
$\frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\,\frac{x}{{\sin x}}\,dx} $

✓

Answer

Correct option: C.

$\frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\,\frac{x}{{\sin x}}\,dx} $

c
$I =$$\int\limits_0^1 {\frac{{{{\tan }^{ - 1}}x}}{x}\,\,dx\,} $ $x =$ $\tan\theta$ ; $dx = \sec^2\theta d\theta$
$I =$ $\int\limits_0^{\frac{\pi }{4}} {\frac{{\theta \,.\,{{\sec }^2}\theta }}{{\tan \theta }}\,d\theta } \,$ $=$ $\int\limits_0^{\frac{\pi }{4}} {\frac{{\theta \,}}{{\sin \theta \,\,\cos \theta }}\,d\theta } \,$ $=$ $\int\limits_0^{\frac{\pi }{4}} {\frac{{2\,\theta \,}}{{\sin 2\theta \,}}\,d\theta } \,$ $(2\theta = y)$
$=$$\frac{1}{2}\,\int\limits_0^{\frac{\pi }{2}} {\frac{y}{{\sin y}}\,\,dy} \,$ $=$ $\frac{1}{2}\,\int\limits_0^{\frac{\pi }{2}} {\frac{x}{{\sin x}}\,\,dx} \,$

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