int_0^{frac{pi}{4}}(sqrt{tan x}+sqrt{cot x}) | vedclass

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(D) Let $I = \int_0^{\frac{\pi}{4}}(\sqrt{	an x}+\sqrt{\cot x}) d x$. We can rewrite the integrand as: $I = \int_0^{\frac{\pi}{4}} \left( \frac{\sqrt

Vedclass · Accepted Answer

$\frac{\pi}{\sqrt{2}}$

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(D) Let $I = \int_0^{\frac{\pi}{4}}(\sqrt{	an x}+\sqrt{\cot x}) d x$. We can rewrite the integrand as: $I = \int_0^{\frac{\pi}{4}} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} ight) d x$ $I = \int_0^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x$ Multiply the numerator and denominator by $\sqrt{2}$: $I = \sqrt{2} \int_0^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} d x = \sqrt{2} \int_0^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} d x$ Using the identity $\sin 2x = 1 - (\sin x - \cos x)^2$: $I = \sqrt{2} \int_0^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} d x$ Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) d x$. When $x = 0$,$u = -1$. When $x = \frac{\pi}{4}$,$u = 0$. $I = \sqrt{2} \int_{-1}^0 \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} [\arcsin u]_{-1}^0$ $I = \sqrt{2} (\arcsin 0 - \arcsin(-1)) = \sqrt{2} (0 - (-\frac{\pi}{2})) = \frac{\sqrt{2} \pi}{2} = \frac{\pi}{\sqrt{2}}$.

$\int_0^{\frac{\pi}{4}}(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

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