The value of the integral int_0^{pi /4} frac{ | vedclass

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Question

(B) Let $I = \int_0^{\pi /4} \frac{\sqrt{	an x}}{\sin x \cos x} \, dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int_0^{\pi /4} \fra

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$2$

Vedclass · Answer

(B) Let $I = \int_0^{\pi /4} \frac{\sqrt{	an x}}{\sin x \cos x} \, dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int_0^{\pi /4} \frac{\sqrt{	an x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \, dx = \int_0^{\pi /4} \frac{\sqrt{	an x} \sec^2 x}{	an x} \, dx = \int_0^{\pi /4} \frac{\sec^2 x}{\sqrt{	an x}} \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. When $x = 0$,$t = 0$. When $x = \pi / 4$,$t = 1$. Substituting these into the integral: $I = \int_0^1 \frac{1}{\sqrt{t}} \, dt = \int_0^1 t^{-1/2} \, dt$. Integrating,we get: $I = [2t^{1/2}]_0^1 = 2(1) - 2(0) = 2$.

The value of the integral $\int_0^{\pi /4} \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx$ equals

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