Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{e^{\sin x}+1}$ is equal to
A
$0$
B
1
C
$-\frac{\pi}{2}$
✓
$\frac{\pi}{2}$
✓
Answer
Correct option: D.
$\frac{\pi}{2}$
(D) Let $I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{ e ^{\sin x}+1}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{d x}{ e ^{\sin x}\left(1+ e ^{-\sin x}\right)}$ $\therefore I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{ e ^{-\sin x}}{1+ e ^{-\sin x}} d x$ ...(i) Also, $I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{ e ^{\sin x}+1} d x$ $=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{ e ^{\sin (-x)}+1} d x$ $\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$ $\therefore \quad I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{e^{-\sin x}+1} d x$ ...(ii) Adding (i) and (ii), we get $2 I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d x$ $\Rightarrow 2 I =[x]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \Rightarrow I =\frac{\pi}{2}$
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