The integral [_0^ / 4 (136 x)/(3 x+5 x) ~d x] is equal to :

Question

The integral $$\int_\limits0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x$$ is equal to :

Accepted Answer

$$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$$

$$\begin{aligned}
& \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \
& \[\begin{gathered} 3 A-5 B=1 \ 5 A+3 B=0 \end{gathered}\end{aligned}$$>A=\frac{3}{34} \quad B=\frac{-5}{34}
\]

$$\begin{aligned}
& \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \
& \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \
& 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \
& 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \
& \Rightarrow 3 \pi-50 \ln 2+20 \ln 5
\end{aligned}$$

Solutions · 1