Definite Integration Question 445
Question: The value of $ \int_0^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt+\int_0^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}dt}} $ is
[MP PET 2001; Orissa JEE 2005]
Options:
A) $ \frac{\pi }{2} $
B) 1
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ I=\int_0^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}dt+\int_0^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}}dt} $
Putting $ t={{\sin }^{2}}u $ in the first integral and $ t={{\cos }^{2}}v $ in the second integral, we have $ I=\int_0^{x}{u\sin 2udu-\int _{\pi /2}^{x}{v\sin 2vdv}} $
$ =\int_0^{\pi /2}{u\sin 2udu+\int _{\pi /2}^{x}{u\sin 2udu-\int _{\pi /2}^{x}{v\sin 2vdv}}} $
$ I=\int_0^{\pi /2}{u\sin 2udu=( \frac{-u\cos 2u}{2} )}_0^{\pi /2}+\frac{1}{2}\int_0^{\pi /2}{\cos 2udu} $
$ =( \frac{-u\cos 2u}{2} )_0^{\pi /2}+\frac{1}{4}(\sin 2u)_0^{\pi /2}=\frac{\pi }{4} $ .
BETA
