The integral $$\int_0^{\pi/4} \frac{136\sin x}{3\sin x + 5\cos x} dx$$ is equal to :

We need $$I = \int_0^{\pi/4} \frac{136\sin x}{3\sin x + 5\cos x}dx$$.

Method: Write $$136\sin x = A(3\sin x + 5\cos x) + B \cdot \frac{d}{dx}(3\sin x + 5\cos x)$$.

$$\frac{d}{dx}(3\sin x + 5\cos x) = 3\cos x - 5\sin x$$.

$$ 136\sin x = A(3\sin x + 5\cos x) + B(3\cos x - 5\sin x) $$ $$ 136\sin x = (3A - 5B)\sin x + (5A + 3B)\cos x $$

Comparing: $$3A - 5B = 136$$ and $$5A + 3B = 0$$.

From the second: $$A = -3B/5$$. Substituting: $$-9B/5 - 5B = 136 \Rightarrow -34B/5 = 136 \Rightarrow B = -20$$.

$$A = -3(-20)/5 = 12$$.

$$ I = 12\int_0^{\pi/4} dx + (-20)\int_0^{\pi/4} \frac{3\cos x - 5\sin x}{3\sin x + 5\cos x}dx $$ $$ = 12 \cdot \frac{\pi}{4} - 20[\ln|3\sin x + 5\cos x|]_0^{\pi/4} $$ $$ = 3\pi - 20\left[\ln\left(3 \cdot \frac{1}{\sqrt{2}} + 5 \cdot \frac{1}{\sqrt{2}}\right) - \ln(5)\right] $$ $$ = 3\pi - 20\left[\ln\frac{8}{\sqrt{2}} - \ln 5\right] = 3\pi - 20\left[\ln\frac{8}{5\sqrt{2}}\right] $$ $$ = 3\pi - 20\ln\frac{8}{5\sqrt{2}} = 3\pi - 20\ln\frac{4\sqrt{2}}{5} $$ $$ = 3\pi - 20[\ln 4 + \ln\sqrt{2} - \ln 5] = 3\pi - 20[2\ln 2 + \frac{1}{2}\ln 2 - \ln 5] $$ $$ = 3\pi - 20 \cdot \frac{5}{2}\ln 2 + 20\ln 5 = 3\pi - 50\ln 2 + 20\ln 5 $$

The correct answer is Option (1): $$3\pi - 50\log_e 2 + 20\log_e 5$$.