Integral Calculus Question 236
Question: $ \int_{{}}^{{}}{x\sqrt{\frac{1-x^{2}}{1+x^{2}}}}\ dx= $
Options:
A) $ \frac{1}{2}[{{\sin }^{-1}}x^{2}+\sqrt{1-x^{4}}]+c $
B) $ \frac{1}{2}[{{\sin }^{-1}}x^{2}+\sqrt{1-x^{2}}]+c $
C) $ {{\sin }^{-1}}x^{2}+\sqrt{1-x^{4}}+c $
D) $ {{\sin }^{-1}}x^{2}+\sqrt{1-x^{2}}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{x\sqrt{\frac{1-x^{2}}{1+x^{2}}}}dx=\int_{{}}^{{}}{\frac{x.,(1-x^{2})}{\sqrt{1-x^{4}}}}dx $ {Multiplying $ N^{r} $ and $ D^{r} $ by $ {{(1-x^{2})}^{1/2}}} $
$ =\int_{{}}^{{}}{\frac{x}{\sqrt{1-x^{4}}}},dx-\int_{{}}^{{}}{\frac{x^{3}}{\sqrt{1-x^{4}}}}dx $ $ =\frac{1}{2}[{{\sin }^{-1}}(x^{2})+\sqrt{1-x^{4}}]+c $ . (By putting $ x^{2}=t $ and $ \sqrt{1-x^{4}}=\sqrt{t} $ respectively)
BETA
