SATHEE: Definite Integration Question 36

Definite Integration Question 36

Question 36

The value of $\int_{0}^{π / 2} \frac{d x}{1 + \tan^{3} x}$ is

(1993, 1M) (a) 0 (b) 1 (c) $π / 2$ (d) $π / 4$

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Solution:

Let $I = \int_{0}^{π / 2} \frac{1}{1 + \tan^{3} x} d x = \int_{0}^{π / 2} \frac{1}{1 + \frac{\sin^{3} x}{\cos^{3} x}} d x$

$\Rightarrow I = \int_{0}^{π / 2} \frac{\cos^{3} x}{\cos^{3} x + \sin^{3} x} d x$

$\Rightarrow I = \int_{0}^{π / 2} \frac{\cos^{3} \frac{π}{2} - x}{\cos^{3} \frac{π}{2} - x + \sin^{3} \frac{π}{2} - x} d x$

$\Rightarrow I = \int_{0}^{π / 2} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} d x$

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

$2 I = \int_{0}^{π / 2} 1 d x \Rightarrow 2 I = [x]_{0}^{π / 2} = π / 2 \Rightarrow I = π / 4$

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Definite Integration Question 36

Question 36

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