If int frac{1 + sqrt{tan x}}{sin 2x} dx = A l | vedclass

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(C) Let $I = \int \frac{1 + \sqrt{	an x}}{\sin 2x} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{1 + \sqrt{	an x}}{2

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(C) Let $I = \int \frac{1 + \sqrt{	an x}}{\sin 2x} dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{1 + \sqrt{	an x}}{2 \sin x \cos x} dx = \int \frac{1}{2 \sin x \cos x} dx + \int \frac{\sqrt{	an x}}{2 \sin x \cos x} dx$. Dividing the numerator and denominator by $\cos^2 x$ in the first integral: $\int \frac{\sec^2 x}{2 	an x} dx = \frac{1}{2} \ln |	an x|$. For the second integral: $\int \frac{\sqrt{	an x}}{2 \sin x \cos x} dx = \int \frac{\sqrt{	an x}}{2 	an x \cos^2 x} dx = \int \frac{\sec^2 x}{2 \sqrt{	an x}} dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. $\int \frac{1}{2 \sqrt{u}} du = \sqrt{u} = \sqrt{	an x}$. Thus,$I = \frac{1}{2} \ln |	an x| + \sqrt{	an x} + C$. Comparing with $A \log 	an x + B \sqrt{	an x} + C$,we get $A = \frac{1}{2}$ and $B = 1$. Therefore,$4A - B = 4(\frac{1}{2}) - 1 = 2 - 1 = 1$.

If $\int \frac{1 + \sqrt{\tan x}}{\sin 2x} dx = A \log \tan x + B \sqrt{\tan x} + C$ then $4A - B =$

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