JEE Main 2024 — 5 April, Shift 1 — Question 9

$I=\int_{0}^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$

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

$136=3 A-5 B$

$0=5 \mathrm{~A}+3 \mathrm{~B}$

$3 B=-5 A \Rightarrow B=-\frac{5}{3} A$

$136=3 \mathrm{~A}-5\left(-\frac{5}{3} \mathrm{~A}\right)$

$136=3 \mathrm{~A}+\frac{25}{3} \mathrm{~A}$

$136=\frac{34 \mathrm{~A}}{3}$

$\Rightarrow \mathrm{A}=\frac{136 \times 3}{34}=12$

$B=\frac{-5}{3}(12)=-20$ $\mathrm{I}=\int_{0}^{\pi / 4} \frac{\mathrm{~A}(3 \sin \mathrm{x}+5 \cos \mathrm{x})}{3 \sin \mathrm{x}+5 \cos \mathrm{x}}+\int_{0}^{\pi / 4} \frac{\mathrm{~B}(3 \cos \mathrm{x}-5 \sin \mathrm{x})}{3 \sin \mathrm{x}+5 \cos \mathrm{x}}$

$=\mathrm{A}(\mathrm{x})_{0}^{\pi / 4}+\mathrm{B}[\ln (3 \sin \mathrm{x}+5 \cos \mathrm{x})]_{0}^{\pi / 4}$

$=12\left(\frac{\pi}{4}\right)-20 \ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5)$ $=3 \pi-20 \ln 4 \sqrt{2}+20 \ell \operatorname{n} 5$

$=3 \pi-20 \times \frac{5}{2} \ln 2+20 \ell \ln 5$

$=3 \pi-50 \ell \operatorname{n} 2+20 \ell \operatorname{n} 5$