The integral ∫_0^π / 4 136 sin x 3 sin x+5 cos x d x is equal to :

$\displaystyle \begin{aligned} & I=\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x \\ & 136 \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & 136=3 A-5 B...(i) \\ & 0=5 A+3 B...(ii)\end{aligned}$​I=∫0π/4​3sinx+5cosx136sinx​dx136sinx=A(3sinx+5cosx)+B(3cosx−5sinx)136=3A−5B...(i)0=5A+3B...(ii)​ $\begin{aligned} & 3 \mathrm{~B}=-5 \mathrm{~A} \Rightarrow \mathrm{B}=-\frac{5}{3} \mathrm{~A} \\ & 136=3 \mathrm{~A}-5\left(-\frac{5}{3} \mathrm{~A}\right) \\ & 136=3 \mathrm{~A}+\frac{25}{3} \mathrm{~A} \\ & 136=\frac{34 \mathrm{~A}}{3} \\ & \Rightarrow \mathrm{A}=\frac{136 \times 3}{34}=12 \\ & \mathrm{~B}=\frac{-5}{3}(12)=-20\end{aligned}$​3 B=−5 A⇒B=−35​ A136=3 A−5(−35​ A)136=3 A+325​ A136=334 A​⇒A=34136×3​=12 B=3−5​(12)=−20​ $\displaystyle \begin{aligned} & I=\int_0^{\pi / 4} \frac{A(3 \sin x+5 \cos x)}{3 \sin x+5 \cos x}+\int_0^{\pi / 4} \frac{B(3 \cos x-5 \sin x)}{3 \sin x+5 \cos x} \\ & =A(x)_0^{\pi / 4}+B[\ln (3 \sin x+5 \cos x)]_0^{\pi / 4} \\ & =12\left(\frac{\pi}{4}\right)-20 \ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5) \\ & =3 \pi-20 \ln 4 \sqrt{2}+20 \ln 5 \\ & =3 \pi-20 \times \frac{5}{2} \ln 2+20 \ln 5 \\ & =3 \pi-50 \ln 2+20 \ln 5\end{aligned}$​I=∫0π/4​3sinx+5cosxA(3sinx+5cosx)​+∫0π/4​3sinx+5cosxB(3cosx−5sinx)​=A(x)0π/4​+B[ln(3sinx+5cosx)]0π/4​=12(4π​)−20ln(2​3​+2​5​)−ln(0+5)=3π−20ln42​+20ln5=3π−20×25​ln2+20ln5=3π−50ln2+20ln5​