Indefinite Integration 1 Question 30
31. $\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\cos x}$ is equal to
$(1999,2 M)$
(a) 2
(b) -2
(c) $\frac{1}{2}$
(d) $-\frac{1}{2}$
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Solution:
- Let
$$ I=\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\cos x} $$
$$ \begin{aligned} \Rightarrow \quad I & =\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1+\cos (\pi-x)} \\ I & =\int _{\pi / 4}^{3 \pi / 4} \frac{d x}{1-\cos x} \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int _{\pi / 4}^{3 \pi / 4} \frac{1}{1+\cos x}+\frac{1}{1-\cos x} d x \\ \Rightarrow \quad 2 I & =\int _{\pi / 4}^{3 \pi / 4} \frac{2}{1-\cos ^{2} x} d x \\ \Rightarrow \quad I & =\int _{\pi / 4}^{3 \pi / 4} \operatorname{cosec}^{2} x d x=[-\cot x] _{\pi / 4}^{3 \pi / 4} \\ & =-\cot \frac{3 \pi}{4}+\cot \frac{\pi}{4}=-(-1)+1=2 \end{aligned} $$
