JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 11)
The integral $$\int_\limits0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x$$ is equal to :
$$3 \pi-50 \log _e 2+20 \log _e 5$$
$$3 \pi-25 \log _e 2+10 \log _e 5$$
$$3 \pi-10 \log _{\mathrm{e}}(2 \sqrt{2})+10 \log _{\mathrm{e}} 5$$
$$3 \pi-30 \log _e 2+20 \log _e 5$$
Explanation
$$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$$
$$\begin{aligned} & \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & \begin{array}{l} 3 A-5 B=1 \\ 5 A+3 B=0 \end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34} \end{aligned}$$
$$\begin{aligned} & \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\ & \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\ & 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\ & 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\ & \Rightarrow 3 \pi-50 \ln 2+20 \ln 5 \end{aligned}$$
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