int_{0}^{pi / 4} frac{sin x+cos x}{3+sin 2 x} | vedclass

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(A) Let $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$. So,the in

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$\frac{1}{4} \log 3$

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(A) Let $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$. So,the integral becomes $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3 + 1 - (\sin x - \cos x)^2} d x = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{4 - (\sin x - \cos x)^2} d x$. Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$. When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$. Thus,$I = \int_{-1}^{0} \frac{dt}{4 - t^2} = \int_{-1}^{0} \frac{dt}{2^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we get: $I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^{0} = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$. $I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

$\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is

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