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\int_{0}^{\pi /2} \frac{1}{1+\sqrt{tanx}} dx

  1. \pi

  2. \pi/2

  3. \pi/4

  4. 3\pi/2

Answers (1)

I = \int_{0}^{\pi /2} \frac{1}{1+\sqrt{\tan x}}dx = \int_{0}^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx...............................(1)

We also know that, \int_{a}^{b} f(x)dx = \int_{a}^{b} f(a+b-x)dx

Using this property in equation (1) , I = \int_{0}^{\pi /2} \frac{\sqrt{\cos (\frac{\pi}{2} -x)}}{\sqrt{\cos (\frac{\pi}{2} -x)}+\sqrt{\sin (\frac{\pi}{2} -x)}}.dx = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx .....................(2)

Adding equation (1) and (2), 2I = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx + \int_{0}^{\pi /2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}.dx

\Rightarrow 2I = \int_{0}^{\pi /2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx = \int_{0}^{\pi /2} 1.dx

2I = \frac{\pi}{2} - 0 \Rightarrow I = \frac{\pi}{4}

Option (3) is correct

Posted by

Abhishek Sahu

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