GATE 2023 AG – Question 13
Evaluate $I=\int_0^{\pi/2}(\sin x+\cos x)^2/\sqrt{1+\sin2x}\,dx$.
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Correct answer: (C) 2
Explanation
On the integration interval, sin x+cos x is positive and $(\sin x+\cos x)^2=1+\sin2x$. Therefore the denominator is sin x+cos x and the integrand reduces to that same sum. Integrating gives $[-\cos x+\sin x]_0^{\pi/2}=1-(-1)=\mathbf2$,**C**.