Prove that $\int_{0}^{\frac{π}{4}} \sqrt{\tan x} + \sqrt{\cot x} dx = \sqrt{2} . \frac{π}{2}$

Solution

$\int_{0}^{\frac{π}{4}} \sqrt{\tan x} + \sqrt{\cot x} dx = \int_{0}^{\frac{π}{4}} (\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}) dx = \int_{0}^{\frac{π}{4}} (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}) dx = \sqrt{2} \int_{0}^{\frac{π}{4}} (\frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}}) dx = \sqrt{2} \int_{0}^{\frac{π}{4}} (\frac{\sin x + \cos x}{\sqrt{1 - \sin x - \cos x^{2}}}) dx Put \sin x - \cos x = t \Rightarrow (\cos x + \sin x) dx = dt$

If x = 0, t = 0 - 1 = - 1

$and if x = \frac{π}{4}, t = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 ∴ \int_{0}^{\frac{π}{4}} \sqrt{\tan x} + \sqrt{\cot x} dx = \sqrt{2} \int_{- 1}^{0} \frac{dt}{\sqrt{1 - t^{2}}} = \sqrt{2} {[\sin^{- 1} t]}_{- 1}^{0} = \sqrt{2} [\sin^{- 1} 0 - \sin^{- 1} (- 1)] = \sqrt{2} [0 + \frac{π}{2}] = \sqrt{2} \times \frac{π}{2} .$

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Prove that $\int_{0}^{\frac{π}{4}} \sqrt{\tan x} + \sqrt{\cot x} dx = \sqrt{2} . \frac{π}{2}$

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Prove that $\int_{0}^{\frac{π}{4}} \sqrt{\tan x} + \sqrt{\cot x} dx = \sqrt{2} . \frac{π}{2}$