Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}& I={\int }_0^{\pi /4}\frac{136\sin x}{3\sin x+5\cos x}dx\\ & 136\sin x=A(3\sin x+5\cos x)+B(3\cos x-5\sin x)\\ & 136=3A-5B...(i)\\ & 0=5A+3B...(ii)\end{array}$ $\begin{array}{rl}& 3\mathrm{B}=-5\mathrm{A}\Rightarrow \mathrm{B}=-\frac{5}{3}\mathrm{A}\\ & 136=3\mathrm{A}-5\left(-\frac{5}{3}\mathrm{A}\right)\\ & 136=3\mathrm{A}+\frac{25}{3}\mathrm{A}\\ & 136=\frac{34\mathrm{A}}{3}\\ & \Rightarrow \mathrm{A}=\frac{136\times 3}{34}=12\\ & \mathrm{B}=\frac{-5}{3}(12)=-20\end{array}$ $\begin{array}{rl}& I={\int }_0^{\pi /4}\frac{A(3\sin x+5\cos x)}{3\sin x+5\cos x}+{\int }_0^{\pi /4}\frac{B(3\cos x-5\sin x)}{3\sin x+5\cos x}\\ & =A(x{)}_0^{\pi /4}+B[\ln (3\sin x+5\cos x){]}_0^{\pi /4}\\ & =12\left(\frac{\pi }{4}\right)-20\ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5)\\ & =3\pi -20\ln 4\sqrt{2}+20\ln 5\\ & =3\pi -20\times \frac{5}{2}\ln 2+20\ln 5\\ & =3\pi -50\ln 2+20\ln 5\end{array}$