SATHEE: Definite Integration Question 30

Definite Integration Question 30

Question 30

$\int_{π / 4}^{3 π / 4} \frac{d x}{1 + \cos x}$ is equal to

$(1999, 2 M)$ (a) 2 (b) -2 (c) $\frac{1}{2}$ (d) $- \frac{1}{2}$

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Solution:

$I = \int_{π / 4}^{3 π / 4} \frac{d x}{1 + \cos x}$

$\begin{aligned} \Rightarrow I & = \int_{π / 4}^{3 π / 4} \frac{d x}{1 + \cos (π - x)} I & = \int_{π / 4}^{3 π / 4} \frac{d x}{1 - \cos x} \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & = \int_{π / 4}^{3 π / 4} \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} d x \Rightarrow 2 I & = \int_{π / 4}^{3 π / 4} \frac{2}{1 - \cos^{2} x} d x \Rightarrow I & = \int_{π / 4}^{3 π / 4} {cosec}^{2} x d x = [- \cot x]_{π / 4}^{3 π / 4} & = - \cot \frac{3 π}{4} + \cot \frac{π}{4} = - (- 1) + 1 = 2 \end{aligned}$

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Definite Integration Question 30

Question 30

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