Indefinite Integration 1 Question 16
17. $\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\cos x}$ is equal to
(a) -2
(b) 2
(c) 4
(d) -1
(2017 Main)
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Solution:
- Let $I=\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\cos x}=\int _{\pi / 4}^{3 \pi / 4} \frac{1-\cos x}{1-\cos ^{2} x} d x$
$$ \begin{aligned} & =\int _{\pi / 4}^{3 \pi / 4} \frac{1-\cos x}{\sin ^{2} x} d x \\ & =\int _{\pi / 4}^{3 \pi / 4}\left(\operatorname{cosec}^{2} x-\operatorname{cosec} x \cot x\right) d x \\ & =[-\cot x+\operatorname{cosec} x] _{\pi / 4}^{3 \pi / 4} \\ & =[(1+\sqrt{2})-(-1+\sqrt{2})]=2 \end{aligned} $$
