If f(x) = int frac{sin 2x + 2 cos x}{4 sin^2 | vedclass

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(C) Given $f(x) = \int \frac{2 \sin x \cos x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$. Factor the denominator: $4 \sin^2 x + 5 \sin x + 1 = (4 \s

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$\frac{1}{2} \log 3$

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(C) Given $f(x) = \int \frac{2 \sin x \cos x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$. Factor the denominator: $4 \sin^2 x + 5 \sin x + 1 = (4 \sin x + 1)(\sin x + 1)$. So, $f(x) = \int \frac{2 \cos x (\sin x + 1)}{(4 \sin x + 1)(\sin x + 1)} \, dx = \int \frac{2 \cos x}{4 \sin x + 1} \, dx$. Let $u = 4 \sin x + 1$, then $du = 4 \cos x \, dx$, which implies $\cos x \, dx = \frac{du}{4}$. $f(x) = \int \frac{2}{u} \cdot \frac{du}{4} = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |4 \sin x + 1| + C$. Given $f(0) = 0$, we have $0 = \frac{1}{2} \ln |4 \sin(0) + 1| + C \implies 0 = \frac{1}{2} \ln(1) + C \implies C = 0$. Thus, $f(x) = \frac{1}{2} \ln |4 \sin x + 1|$. Now, $f\left(\frac{\pi}{6}\right) = \frac{1}{2} \ln |4 \sin(\frac{\pi}{6}) + 1| = \frac{1}{2} \ln |4(\frac{1}{2}) + 1| = \frac{1}{2} \ln(3)$.

If $f(x) = \int \frac{\sin 2x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$ and $f(0) = 0$, then $f\left(\frac{\pi}{6}\right) =$

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