Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^x \frac{x \tan x}{\sec x+\cos x} d x=$
✓
$\frac{\pi^2}{4}$
B
$\frac{\pi^2}{2}$
C
$\frac{3 \pi^2}{2}$
D
$\frac{\pi^2}{3}$
✓
Answer
Correct option: A.
$\frac{\pi^2}{4}$
(A) $\text { Let } I=\int_0^\pi \frac{x \tan x}{\sec x+\cos x} d x$ ...(i) $\therefore \quad I=\int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\cos x} d x$ ...(ii) $\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$ Adding (i) and (ii), we get $2 I =\pi \int_0^\pi \frac{\tan x}{\sec x+\cos x} d x$ $\Rightarrow I =\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$ Put $\cos x= t \Rightarrow \sin x d x=- dt$ $\therefore \quad I=-\frac{\pi}{2} \int_1^{-1} \frac{d t}{1+t^2}=-\frac{\pi}{2}\left[\tan ^{-1} t\right]_1^{-1}$ $=\left(-\frac{\pi}{2}\right)\left(-\frac{\pi}{2}\right)=\frac{\pi^2}{4}$
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