Integral Calculus Question 495
Question: $ \int_{{}}^{{}}{\frac{x}{\sqrt{4-x^{4}}}dx}= $
[Roorkee 1976]
Options:
A) $ {{\cos }^{-1}}\frac{x^{2}}{2} $
B) $ \frac{1}{2}{{\cos }^{-1}}\frac{x^{2}}{2} $
C) $ {{\sin }^{-1}}\frac{x^{2}}{2} $
D) $ \frac{1}{2}{{\sin }^{-1}}\frac{x^{2}}{2} $
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{x}{\sqrt{4-x^{4}}}},dx=\int_{{}}^{{}}{\frac{x}{\sqrt{2^{2}-{{(x^{2})}^{2}}}}},dx $ Putting $ x^{2}=t\Rightarrow 2x,dx=dt, $ we get the required integral $ =\frac{1}{2}{{\sin }^{-1}}\frac{x^{2}}{2} $ .
BETA
