Integral Calculus Question 383
Question: $ \int_{{}}^{{}}{{{\sin }^{-1}}x\ dx} $ is equal to
[MP PET 2004]
Options:
A) $ \frac{1}{\sqrt{1-x^{2}}}+c $
B) $ x{{\sin }^{-1}}x-\sqrt{1-x^{2}}+c $
C) $ {{\cos }^{-1}}x+c $
D) $ x{{\sin }^{-1}}x+\sqrt{1-x^{2}}+c $
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Answer:
Correct Answer: D
Solution:
$ I=\int_{{}}^{{}}{{{\sin }^{-1}}x}.1,dx,dx $ $ I={{\sin }^{-1}}x.x-\int_{{}}^{{}}{\frac{1}{\sqrt{1-x^{2}}}},.,x,dx $ Put $ 1-x^{2}=t^{2}\Rightarrow -2xdx=2tdt $ in the second integral and solve it, therefore $ I=x{{\sin }^{-1}}x.+\sqrt{1-x^{2}}+c $ .
BETA
