Integral Calculus Question 111
Question: $ \int_{{}}^{{}}{x{{\sin }^{-1}}x\ dx}= $
[MP PET 1991]
Options:
A) $ ( \frac{x^{2}}{2}-\frac{1}{4} ){{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-x^{2}}+c $
B) $ ( \frac{x^{2}}{2}+\frac{1}{4} ){{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-x^{2}}+c $
C) $ ( \frac{x^{2}}{2}-\frac{1}{4} ){{\sin }^{-1}}x-\frac{x}{4}\sqrt{1-x^{2}}+c $
D) $ ( \frac{x^{2}}{2}+\frac{1}{4} ){{\sin }^{-1}}x-\frac{x}{4}\sqrt{1-x^{2}}+c $
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Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{x{{\sin }^{-1}}xdx=\frac{x^{2}}{2}{{\sin }^{-1}}x-\int_{{}}^{{}}{\frac{1}{\sqrt{1-x^{2}}}.\frac{x^{2}}{2}dx+c}} $ $ =\frac{x^{2}}{2}{{\sin }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{-\frac{(1-x^{2})+1}{\sqrt{1-x^{2}}}}dx+c $
$ =\frac{x^{2}}{2}{{\sin }^{-1}}x+\frac{1}{2}\int_{{}}^{{}}{\sqrt{1-x^{2}}dx-\frac{1}{2}\int_{{}}^{{}}{\frac{1}{\sqrt{1-x^{2}}}dx+c}} $ $ =\frac{x^{2}}{2}{{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-x^{2}}+\frac{1}{4}{{\sin }^{-1}}x-\frac{1}{2}{{\sin }^{-1}}x+c $
$ =\frac{x^{2}}{2}{{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-x^{2}}-\frac{1}{4}{{\sin }^{-1}}x $
$ =( \frac{x^{2}}{2}-\frac{1}{4} ){{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-x^{2}}+c $ .
BETA
