The value of $\mathrm{I}=\int_{0}^{\pi / 2} \frac{(\sin \mathrm{x}+\cos \mathrm{x})^{2}}{\sqrt{1+\sin 2 \mathrm{x}}} \mathrm{dx}$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Definite Integrals

The value of

I = \int_{0}^{π /2} \frac{( s i n x + c o s x ) ^{2}}{1 + s i n 2 x} dx

Answer: (C)

Since

(sin x + cos x)^{2} = 1 + sin 2 x

, the integrand simplifies to

\frac{1 + s i n 2 x}{1 + s i n 2 x} = 1 + sin 2 x = ∣ sin x + cos x ∣

, which is just

sin x + cos x

[0, \frac{π}{2}]

where it stays positive. Therefore

I = \int_{0}^{π /2} (sin x + cos x) d x = [sin x - cos x]_{0}^{π /2} = (1 - 0) - (0 - 1) = 2

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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