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The integral $\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$ is equal to :
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The correct answer is:
$3 \pi-50 \log _e 2+20 \log _e 5$
$\begin{aligned} & I=\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x \\ & 136 \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & 136=3 A-5 B...(i) \\ & 0=5 A+3 B...(ii)\end{aligned}$
$\begin{aligned} & 3 \mathrm{~B}=-5 \mathrm{~A} \Rightarrow \mathrm{B}=-\frac{5}{3} \mathrm{~A} \\ & 136=3 \mathrm{~A}-5\left(-\frac{5}{3} \mathrm{~A}\right) \\ & 136=3 \mathrm{~A}+\frac{25}{3} \mathrm{~A} \\ & 136=\frac{34 \mathrm{~A}}{3} \\ & \Rightarrow \mathrm{A}=\frac{136 \times 3}{34}=12 \\ & \mathrm{~B}=\frac{-5}{3}(12)=-20\end{aligned}$
$\begin{aligned} & I=\int_0^{\pi / 4} \frac{A(3 \sin x+5 \cos x)}{3 \sin x+5 \cos x}+\int_0^{\pi / 4} \frac{B(3 \cos x-5 \sin x)}{3 \sin x+5 \cos x} \\ & =A(x)_0^{\pi / 4}+B[\ln (3 \sin x+5 \cos x)]_0^{\pi / 4} \\ & =12\left(\frac{\pi}{4}\right)-20 \ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5) \\ & =3 \pi-20 \ln 4 \sqrt{2}+20 \ln 5 \\ & =3 \pi-20 \times \frac{5}{2} \ln 2+20 \ln 5 \\ & =3 \pi-50 \ln 2+20 \ln 5\end{aligned}$
$\begin{aligned} & 3 \mathrm{~B}=-5 \mathrm{~A} \Rightarrow \mathrm{B}=-\frac{5}{3} \mathrm{~A} \\ & 136=3 \mathrm{~A}-5\left(-\frac{5}{3} \mathrm{~A}\right) \\ & 136=3 \mathrm{~A}+\frac{25}{3} \mathrm{~A} \\ & 136=\frac{34 \mathrm{~A}}{3} \\ & \Rightarrow \mathrm{A}=\frac{136 \times 3}{34}=12 \\ & \mathrm{~B}=\frac{-5}{3}(12)=-20\end{aligned}$
$\begin{aligned} & I=\int_0^{\pi / 4} \frac{A(3 \sin x+5 \cos x)}{3 \sin x+5 \cos x}+\int_0^{\pi / 4} \frac{B(3 \cos x-5 \sin x)}{3 \sin x+5 \cos x} \\ & =A(x)_0^{\pi / 4}+B[\ln (3 \sin x+5 \cos x)]_0^{\pi / 4} \\ & =12\left(\frac{\pi}{4}\right)-20 \ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5) \\ & =3 \pi-20 \ln 4 \sqrt{2}+20 \ln 5 \\ & =3 \pi-20 \times \frac{5}{2} \ln 2+20 \ln 5 \\ & =3 \pi-50 \ln 2+20 \ln 5\end{aligned}$
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