int frac{sqrt{tan x}}{sin x cos x} , dx = | vedclass

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Question

(B) Let $I = \int \frac{\sqrt{	an x}}{\sin x \cos x} \, dx$. Multiply the numerator and denominator by $\cos x$: $I = \int \frac{\sqrt{	an x}}{\sin

Vedclass · Accepted Answer

$2\sqrt{	an x} + c$

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{	an x}}{\sin x \cos x} \, dx$. Multiply the numerator and denominator by $\cos x$: $I = \int \frac{\sqrt{	an x}}{\sin x \cos x \cdot \frac{\cos x}{\cos x}} \, dx = \int \frac{\sqrt{	an x}}{	an x \cos^2 x} \, dx = \int \frac{\sec^2 x}{\sqrt{	an x}} \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. Substituting these into the integral: $I = \int \frac{1}{\sqrt{t}} \, dt = \int t^{-1/2} \, dt = \frac{t^{1/2}}{1/2} + c = 2\sqrt{t} + c$. Replacing $t$ with $	an x$,we get $I = 2\sqrt{	an x} + c$.

$\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = $

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