Integral Calculus Question 190
Question: $ \int_{{}}^{{}}{\frac{1+x^{2}}{\sqrt{1-x^{2}}}dx=} $
[IIT 1977]
Options:
A) $ \frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-x^{2}}+c $
B) $ \frac{3}{2}{{\sin }^{-1}}x+\frac{1}{2}x\sqrt{1-x^{2}}+c $
C) $ \frac{3}{2}{{\cos }^{-1}}x-\frac{1}{2}x\sqrt{1-x^{2}}+c $
D) $ \frac{3}{2}{{\cos }^{-1}}x+\frac{1}{2}x\sqrt{1-x^{2}}+c $
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Answer:
Correct Answer: A
Solution:
Put $ x=\sin \theta \Rightarrow dx=\cos \theta ,d\theta , $ then it reduces to $ \int_{{}}^{{}}{(1+{{\sin }^{2}}\theta ),d\theta }=\theta +\frac{1}{2}\int_{{}}^{{}}{(1-\cos 2\theta ),d\theta } $ $ =\frac{3\theta }{2}-\frac{1}{2}\sin \theta \sqrt{1-{{\sin }^{2}}\theta }+c=\frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-x^{2}}+c $ .
BETA
