Integral Calculus Question 448
Question: The value of $ \int_{{}}^{{}}{\frac{dx}{\sqrt{1-x}}} $ is
[Pb. CET 2001]
Options:
A) $ 2\sqrt{1-x}+c $
B) $ -2\sqrt{1-x}+c $
C) $ -{{\sin }^{-1}}\sqrt{x}+c $
D) $ {{\sin }^{-1}}\sqrt{x}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
We have, $ \int_{{}}^{{}}{\frac{dx}{\sqrt{1-x}}} $ or $ I=\int_{{}}^{{}}{{{(1-x)}^{-1/2}}dx} $ $ I=\frac{{{(1-x)}^{\frac{-1}{2}+1}}}{(-1),( -\frac{1}{2}+1 )}+c $
Þ $ I=-2\sqrt{1-x}+c $ .
BETA
