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

MCQ

$\int_0^1 {{{\tan }^{ - 1}}x\,dx = } $

✓
$\frac{\pi }{4} - \frac{1}{2}\log 2$
B
$\pi - \frac{1}{2}\log 2$
C
$\frac{\pi }{4} - \log 2$
D
$\pi - \log 2$

✓

Answer

Correct option: A.

$\frac{\pi }{4} - \frac{1}{2}\log 2$

a
(a) Put $x = \tan \theta $

$\Rightarrow dx = {\sec ^2}\theta \,\,d\theta $

Also as $x = 0,\theta = 0$ and $x = 1,\theta = \frac{\pi }{4}$

Therefore, $\int_0^1 {{{\tan }^{ - 1}}x\,dx = \int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta } } $

$ = \frac{\pi }{4}$$ - \log \sqrt 2 = \frac{\pi }{4} - \frac{1}{2}\log 2$.

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