The value of I = int_0^{pi /2} frac{(sin x + | vedclass

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(C) Given $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$. We know that $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x =

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$2$

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(C) Given $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$. We know that $1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2$. Substituting this into the integral,we get: $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{(\sin x + \cos x)^2}} dx$. Since $\sin x + \cos x > 0$ for $x \in [0, \pi/2]$,$\sqrt{(\sin x + \cos x)^2} = \sin x + \cos x$. Thus,$I = \int_0^{\pi /2} (\sin x + \cos x) dx$. Evaluating the integral: $I = [-\cos x + \sin x]_0^{\pi /2}$. $I = (-\cos(\pi/2) + \sin(\pi/2)) - (-\cos(0) + \sin(0))$. $I = (0 + 1) - (-1 + 0) = 1 + 1 = 2$.

The value of $I = \int_0^{\pi /2} \frac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}} dx$ is

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