A) 3
B) 1
C) 2
D) 0
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{1+\sin 2x}}dx}\]\[=\int_{0}^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{{{(\sin x+\cos x)}^{2}}}}dx}\] \[I=\int_{0}^{\pi /2}{(\sin x+\cos x)dx=(-\cos x+\sin x)_{0}^{\pi /2}}\] \[I=1-(-1)=2\].You need to login to perform this action.
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