Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^\pi \frac{x}{1+\sin x} d x$ is equal to
A
$\frac{\pi}{2}$
✓
$\pi$
C
$\frac{\pi}{2} \log 2$
D
$\pi \log 2$
✓
Answer
Correct option: B.
$\pi$
(B) Let $I=\int_0^\pi \frac{x}{1+\sin x} d x$ ...(i) $\therefore \quad I=\int_0^\pi \frac{\pi-x}{1+\sin x} d x$ ...(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 $2 I =\int_0^\pi \frac{\pi}{1+\sin x} d x$ $\begin{array}{l}=\pi \int_0^\pi \frac{1-\sin x}{(1+\sin x)(1-\sin x)} d x \\ =\pi \int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x=\pi \int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x\end{array}$ $=\pi[\tan x-\sec x]_0^\pi$ $\therefore \quad 2 I =\pi[0-(-1)-(0-1)]=2 \pi$ $\Rightarrow I =\pi$
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