Integral Calculus Question 432
Question: $ \int_{{}}^{{}}{\frac{{{\sin }^{-1}}x}{{{(1-x^{2})}^{3/2}}}\ dx=} $
[AISSE 1983, 87]
Options:
A) $ \frac{x}{\sqrt{1-x^{2}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-x^{2})+c $
B) $ \frac{x}{\sqrt{1-x^{2}}}{{\sin }^{-1}}x-\frac{1}{2}\log (1-x^{2})+c $
C) $ \frac{1}{\sqrt{1-x^{2}}}{{\sin }^{-1}}x-\frac{1}{2}\log (1-x^{2})+c $
D) $ \frac{1}{\sqrt{1-x^{2}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-x^{2})+c $
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Answer:
Correct Answer: A
Solution:
Put $ t={{\sin }^{-1}}x\Rightarrow \sin t=x\Rightarrow \cos t,dt=dx, $ then $ \int_{{}}^{{}}{\frac{{{\sin }^{-1}}x}{{{(1-x^{2})}^{3/2}}},dx}=\int_{{}}^{{}}{t{{\sec }^{2}}t,dt=t\tan t+\log \cos t+c} $ $ ={{\sin }^{-1}}x\tan ({{\sin }^{-1}}x)+\log \cos ({{\sin }^{-1}}x)+c $ $ =\frac{x}{\sqrt{1-x^{2}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-x^{2})+c. $
BETA
