MCQ
$\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx = } $
- ✓$\frac{\pi }{2} - 2\log \sqrt 2 $
- B$\frac{\pi }{2} + 2\log \sqrt 2 $
- C$\frac{\pi }{4} - \log \sqrt 2 $
- D$\frac{\pi }{4} + \log \sqrt 2 $
✓
Answer
Correct option: A.
$\frac{\pi }{2} - 2\log \sqrt 2 $
a
(a) Put $x = \tan \theta ,$
(a) Put $x = \tan \theta ,$
$\therefore $ $dx = {\sec ^2}\theta \,d\theta $
As $x = 1 \Rightarrow \theta = \frac{\pi }{4}$ and
$x = 0 \Rightarrow \theta = 0$, then
$I = 2\int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta = 2[\theta \tan \theta ]_0^{\pi /4} - 2\int_0^{\pi /4} {\tan \theta \,d\theta } } $
$= \frac{\pi }{2} + 2\,[\log \cos x]_0^{\pi /4} = \frac{\pi }{2} - 2\log \sqrt 2 $.
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