SATHEE: Indefinite Integration 1 Question 16

Indefinite Integration 1 Question 16

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

(a) -2

(b) 2

(d) -1

(2017 Main)

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Solution:

Let $I = \int_{π / 4}^{3 π / 4} \frac{d x}{1 + \cos x} = \int_{π / 4}^{3 π / 4} \frac{1 - \cos x}{1 - \cos^{2} x} d x$

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

Indefinite Integration 1 Question 16

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

Solution:

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Indefinite Integration 1 Question 16

17. ∫π/43π/4dx1+cos⁡x is equal to

Solution:

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