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Question: Answered & Verified by Expert
$$
\int_0^x \frac{x}{\sin x}\left(3 \cos ^2 x+2 \sin x+\sin ^3 x-3\right) d x=
$$
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2022 (06 Jul Shift 1)
Options:
  • A $\frac{\pi(5 \pi-12)}{4}$
  • B $\frac{\pi}{2}$
  • C $\frac{\pi}{2}(5 \pi-6)$
  • D $\frac{\pi(5 \pi-12)}{6}$
Solution:
2736 Upvotes Verified Answer
The correct answer is: $\frac{\pi(5 \pi-12)}{4}$
$\begin{aligned} & \text { given } \int_0^\pi \frac{x}{\sin x}\left(3 \cos ^2 x+2 \sin x+\sin ^3 x-3\right) d x \\ & \Rightarrow \int_0^\pi \frac{x}{\sin x}\left(-3 \sin ^2 x+2 \sin x+\sin ^3 x\right) d x \\ & \Rightarrow \int_0^\pi x\left(\sin ^2 x-3 \sin x+2\right) d x\end{aligned}$
Let $I=\int_0^\pi x\left(\sin ^2 x-3 \sin x+2\right) d x ...(1)$
using propert $(a+b-x)$
$$
I=\int_0^\pi(\pi-x)\left(\sin ^2 x-3 \sin x+2\right) d x ...(ii)
$$
adding (1) and (ii)
$$
\begin{aligned}
& 2 I=\int_0^\pi \pi\left(\sin ^2(x)-3 \sin x+2\right) d x \\
& 2 I=\int_0^\pi \pi\left(\frac{1-\cos 2 x}{2}-3 \sin x+2\right) d x \\
& 2 I=\frac{\pi}{2} \int_0^\pi\left(5-\cos ^2 x-6 \sin x\right) d x \\
& I=\frac{\pi}{4}\left[5 x-\frac{\sin 2 x}{2}+6 \cos ^x\right]_0^\pi \\
& =\frac{\pi}{4}\left[\left(5 x \frac{-1}{2} \sin 2 \pi+6 \cos \pi\right)-(0-0+6)\right] \\
& =\frac{\pi}{4}[5 \pi 6-6] \\
& =\frac{\pi}{4}[5 \pi-12]
\end{aligned}
$$

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