int_0^{pi / 4} frac{sin x+cos x}{7+9 sin 2 x} | vedclass

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(D) Let $I = \int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$.We know that $(\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 -

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$\frac{\log 7}{24}$

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(D) Let $I = \int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$.We know that $(\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$.Therefore, $\sin 2x = 1 - (\sin x - \cos x)^2$.Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.When $x = 0$, $t = \sin 0 - \cos 0 = -1$.When $x = \pi/4$, $t = \sin(\pi/4) - \cos(\pi/4) = 0$.Substituting these into the integral:$I = \int_{-1}^0 \frac{dt}{7 + 9(1 - t^2)} = \int_{-1}^0 \frac{dt}{16 - 9t^2}$.$I = \frac{1}{9} \int_{-1}^0 \frac{dt}{(4/3)^2 - t^2}$.Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$, we get:$I = \frac{1}{9} \cdot \frac{1}{2(4/3)} \left[ \log \left| \frac{4/3 + t}{4/3 - t} \right| \right]_{-1}^0 = \frac{1}{24} \left[ \log \left| \frac{4+3t}{4-3t} \right| \right]_{-1}^0$.$I = \frac{1}{24} [\log(1) - \log| (4-3)/(4+3) |] = \frac{1}{24} [0 - \log(1/7)] = \frac{1}{24} \log 7$.

$\int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$ is equal to

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