Integrate the function frac{sqrt{tan x}}{sin | vedclass

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Question

(A) 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} \cdot \co

Vedclass · Accepted Answer

$2\sqrt{	an x} + C$

Vedclass · Answer

(A) Let $I = \int \frac{\sqrt{	an x}}{\sin x \cos x} dx$. Multiply the numerator and denominator by $\cos x$: $I = \int \frac{\sqrt{	an x} \cdot \cos x}{\sin x \cos x \cdot \cos x} dx = \int \frac{\sqrt{	an x}}{	an x \cos^2 x} dx$. Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have: $I = \int \frac{\sec^2 x}{\sqrt{	an x}} dx$. Let $t = 	an x$,then $dt = \sec^2 x dx$. Substituting these into the integral: $I = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt$. Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + C$: $I = \frac{t^{1/2}}{1/2} + C = 2\sqrt{t} + C$. Substituting back $t = 	an x$: $I = 2\sqrt{	an x} + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{\sqrt{\tan x}}{\sin x \cos x}$.

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