Definite Integration Question 315
Question: The value of $ I=\int_0^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{1+\sin 2x}}}dx $ is
[AIEEE 2004]
Options:
A) 3
B) 1
C) 2
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_0^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{1+\sin 2x}}dx} $
$ =\int_0^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{{{(\sin x+\cos x)}^{2}}}}dx} $
$ I=\int_0^{\pi /2}{(\sin x+\cos x)dx=(-\cos x+\sin x)_0^{\pi /2}} $
$ I=1-(-1)=2 $ .
BETA
