If f(x) = sqrt{tan x} and g(x) = sin x cdot c | vedclass

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(A) Given $f(x) = \sqrt{	an x}$ and $g(x) = \sin x \cdot \cos x$. We need to evaluate the integral $I = \int \frac{f(x)}{g(x)} dx$. $I = \int \frac{\

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$2 \sqrt{	an x} + C$

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(A) Given $f(x) = \sqrt{	an x}$ and $g(x) = \sin x \cdot \cos x$. We need to evaluate the integral $I = \int \frac{f(x)}{g(x)} dx$. $I = \int \frac{\sqrt{	an x}}{\sin x \cdot \cos x} dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sqrt{	an x}}{\frac{\sin x \cdot \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} dx = \int \frac{\sqrt{	an x}}{	an x} \cdot \sec^2 x dx$. $I = \int \frac{\sec^2 x}{\sqrt{	an x}} dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. $I = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$. Substituting back $u = 	an x$,we get $I = 2\sqrt{	an x} + C$.

If $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$,then $\int \frac{f(x)}{g(x)} dx$ is equal to (where $C$ is a constant of integration).

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