The value of -0-2 sin x-cos x2-1-sin 2xdx is

I = ∫_0^π/2 (sin x+cos x)^2/√(1+sin 2x)dx = ∫_0^π/2(sin x+cos x)^2/√(cos^2x+sin^2x+2 sin x cos x)dx =∫_0^π/2(sin x+cos x)dx=[ cos x+sin x]^π/2_0=2 ...

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Standard XI

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of $\int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x i s$

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Solution

The correct option is C 2
$I = \int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x = \int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{c o s^{2} x + s i n^{2} x + 2 s i n x c o s x}} d x$
$= \int_{0}^{\frac{π}{2}} (s i n x + c o s x) d x = [- c o s x + s i n x]_{0}^{\frac{π}{2}} = 2$

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The value of $\int_{0}^{\frac{π}{2}} \frac{(s i n x + c o s x)^{2}}{\sqrt{1 + s i n 2 x}} d x i s$