Integral Calculus Question 491
Question: $ \int_{{}}^{{}}{\sqrt{\frac{1-x}{1+x}}}\ dx= $
[IIT 1971]
Options:
A) $ {{\sin }^{-1}}x-\frac{1}{2}\sqrt{1-x^{2}}+c $
B) $ {{\sin }^{-1}}x+\frac{1}{2}\sqrt{1-x^{2}}+c $
C) $ {{\sin }^{-1}}x-\sqrt{1-x^{2}}+c $
D) $ {{\sin }^{-1}}x+\sqrt{1-x^{2}}+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\sqrt{\frac{1-x}{1+x}},dx}=\int_{{}}^{{}}{\frac{1-x}{\sqrt{1-x^{2}}}},dx=\int_{{}}^{{}}{\frac{1}{\sqrt{1-x^{2}}}},dx-\int_{{}}^{{}}{\frac{x,dx}{\sqrt{1-x^{2}}}} $ Now proceed yourself.
BETA
