int_0^{pi /4} [sqrt{tan x} + sqrt{cot x}] , d | vedclass

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(C) Let $I = \int_0^{\pi /4} [\sqrt{	an x} + \sqrt{\cot x}] \, dx$. We can rewrite the integrand as: $I = \int_0^{\pi /4} \left( \sqrt{\frac{\sin x}{

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

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(C) Let $I = \int_0^{\pi /4} [\sqrt{	an x} + \sqrt{\cot x}] \, dx$. We can rewrite the integrand as: $I = \int_0^{\pi /4} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} ight) dx = \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply the numerator and denominator by $\sqrt{2}$: $I = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi /4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$. Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) \, dx$. When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \pi / 4$,$t = \sin(\pi / 4) - \cos(\pi / 4) = 0$. Substituting these into the integral: $I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$. $I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi / 2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

$\int_0^{\pi /4} [\sqrt{\tan x} + \sqrt{\cot x}] \, dx$ equals

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