int_0^{frac{pi}{4}} frac{sin x+cos x}{3+sin 2 | vedclass

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(D) Let $I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x = 0$,$

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

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(D) Let $I = \int_0^{\pi/4} \frac{\sin x + \cos x}{3 + \sin 2x} dx$. Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) dx$. When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$. Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$. Substituting these into the integral: $I = \int_{-1}^0 \frac{dt}{3 + (1 - t^2)} = \int_{-1}^0 \frac{dt}{4 - t^2} = \int_{-1}^0 \frac{dt}{2^2 - t^2}$. Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$: $I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^0 = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$. $I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

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