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Question: Answered & Verified by Expert
$\int_0^\pi \frac{x \tan x}{\sec x+\cos x} \mathrm{~d} x=$
MathematicsDefinite IntegrationMHT CETMHT CET 2023 (10 May Shift 2)
Options:
  • A $\frac{\pi}{8}$
  • B $-\frac{\pi^2}{8}$
  • C $\frac{\pi^2}{4}$
  • D $-\frac{\pi^2}{4}$
Solution:
1544 Upvotes Verified Answer
The correct answer is: $\frac{\pi^2}{4}$
$\begin{aligned} & \text { Let } \mathrm{I}=\int_0^\pi \frac{x \tan x}{\sec x+\cos x} \mathrm{~d} x \quad \ldots \text { (i) } \\ & \therefore \quad \mathrm{I}=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\cos x} \mathrm{~d} x \ldots \text { (ii) } \\ & \ldots\left[\because \int_0^{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x=\int_0^{\mathrm{a}} \mathrm{f}(\mathrm{a}-x) \mathrm{d} x\right]\end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned}
& 2 \mathrm{I}=\pi \int_0^\pi \frac{\tan x}{\sec x+\cos x} \mathrm{~d} x \\
& \Rightarrow \mathrm{I}=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} \mathrm{~d} x
\end{aligned}$
Put $\cos x=\mathrm{t} \Rightarrow \sin x \mathrm{~d} x=-\mathrm{dt}$
$\begin{aligned}
\therefore \quad \mathrm{I} & =-\frac{\pi}{2} \int_1^{-1} \frac{\mathrm{dt}}{1+\mathrm{t}^2} \\
& =-\frac{\pi}{2}\left[\tan ^{-1} \mathrm{t}\right]_1^{-1} \\
& =\left(-\frac{\pi}{2}\right)\left(-\frac{\pi}{2}\right)=\frac{\pi^2}{4}
\end{aligned}$

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