Definite Integration Question 364
Question: $ \int_0^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-x^{2}}}dx=} $
[IIT 1984]
Options:
A) $ \frac{1}{2}+\frac{\sqrt{3}\pi }{12} $
B) $ \frac{1}{2}-\frac{\sqrt{3}\pi }{12} $
C) $ \frac{1}{2}-\frac{\sqrt{3\pi }}{12} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ t={{\sin }^{-1}}x\Rightarrow dt=\frac{1}{\sqrt{1-x^{2}}}dx, $ then $ \int_0^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-x^{2}}}dx=\int_0^{\pi /6}{t\sin tdt}} $
$ =[-t\cos t+\sin t]_0^{\pi /6} $
$ =[ -\frac{\pi }{6}.\frac{\sqrt{3}}{2}+\frac{1}{2} ]=[ \frac{1}{2}-\frac{\sqrt{3}\pi }{12} ] $ .
BETA
