The value of _ / 4 ^ 3 / 4 1+ d is - Vidyadip

MCQ

The value of $\int_{\pi / 4}^{3 \pi / 4} \frac{\phi}{1+\sin \phi} d \phi$ is

✓
$\pi \tan \frac{\pi}{8}$
B
$\log \tan \frac{\pi}{8}$
C
$\tan \frac{\pi}{8}$
D
None of these

✓

Answer

Correct option: A.

$\pi \tan \frac{\pi}{8}$

(A)
Let $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\phi}{1+\sin \phi} d \phi$ ...(i)
$=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi-\phi}{1+\sin (\pi-\phi)} d \phi$
$\ldots \left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi-\phi}{1+\sin \phi} d \phi$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{\pi}{1+\sin (\pi-\phi)} d \phi$
On solving, we get
$2 I =2 \pi(\sqrt{2}-1)$
$\therefore \quad I=\pi(\sqrt{2}-1)=\pi \tan \frac{\pi}{8}$

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