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

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Question

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

Vedclass · Accepted Answer

$2 \sqrt{	an x}+c$, where $c$ is a constant of integration

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(B) Let $I = \int \frac{\sqrt{	an x}}{\sin x \cos x} \,dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sqrt{	an x} / \cos^2 x}{(\sin x \cos x) / \cos^2 x} \,dx = \int \frac{\sqrt{	an x} \cdot \sec^2 x}{	an x} \,dx$. $I = \int \frac{\sec^2 x}{\sqrt{	an x}} \,dx$. Let $u = 	an x$, then $du = \sec^2 x \,dx$. Substituting these into the integral: $I = \int \frac{1}{\sqrt{u}} \,du = \int u^{-1/2} \,du$. Using the power rule $\int u^n \,du = \frac{u^{n+1}}{n+1} + c$: $I = \frac{u^{1/2}}{1/2} + c = 2 \sqrt{u} + c$. Substituting back $u = 	an x$: $I = 2 \sqrt{	an x} + c$.

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

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