MCQ
The integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ is equal to
  • A
    -1
  • B
    -2
  • 2
  • D
    4

Answer

Correct option: C.
2
(C)
Let $I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos x}$ ...(i)
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1+\cos (\pi-x)}$
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
$\therefore \quad I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{d x}{1-\cos x}$ ...(ii)
Adding (i) and (ii), we get
$2 I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{2}{1-\cos ^2 x} d x$
$\therefore \quad I =\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \operatorname{cosec}^2 x d x$
$=-[\cot x]_{\pi / 4}^{3 \pi / 4}=2$

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