Integral Calculus Question 420
Question: $ \int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-x^{2}}}\ }dx= $
[MNR 1978; EAMCET 1982; IIT 1984]
Options:
A) $ x-\sqrt{1-x^{2}}{{\sin }^{-1}}x+c $
B) $ x+\sqrt{1-x^{2}}{{\sin }^{-1}}x+c $
C) $ \sqrt{1-x^{2}}{{\sin }^{-1}}x-x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Putting $ {{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-x^{2}}},dx=dt, $ we get $ \int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-x^{2}}},dx=}\int_{{}}^{{}}{t\sin t,dt=-t\cos t+\sin t+c} $ $ =-{{\sin }^{-1}}x\cos ({{\sin }^{-1}}x)+\sin ({{\sin }^{-1}}x)+c $ $ =x-{{\sin }^{-1}}x\sqrt{1-x^{2}}+c. $
BETA
