If int frac{3}{2 cos ^3 x sqrt{2 sin 2 x}} d | vedclass

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(D) Given the integral $I = \int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x$. Using $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$,we have $\sqrt{2 \sin 2

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$\frac{5}{2}$

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(D) Given the integral $I = \int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x$. Using $\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$,we have $\sqrt{2 \sin 2x} = \sqrt{\frac{4 	an x}{1 + 	an^2 x}} = \frac{2 \sqrt{	an x}}{\sec x}$. Substituting this into the integral: $I = \int \frac{3}{2 \cos^3 x \cdot \frac{2 \sqrt{	an x}}{\sec x}} d x = \int \frac{3 \sec^2 x}{4 \sqrt{	an x}} d x$. Since $\sec^2 x = 1 + 	an^2 x$,we rewrite the integral as: $I = \int \frac{3(1 + 	an^2 x)}{4 \sqrt{	an x}} \sec^2 x d x$. Let $t = 	an x$,then $dt = \sec^2 x d x$. $I = \frac{3}{4} \int (t^{-1/2} + t^{3/2}) dt = \frac{3}{4} [2t^{1/2} + \frac{2}{5} t^{5/2}] + c = \frac{3}{2} (	an x)^{1/2} + \frac{3}{10} (	an x)^{5/2} + c$. Comparing this with the given form $\frac{3}{2}(	an x)^B + \frac{1}{10}(	an x)^A + c$,we note that the coefficient of $(	an x)^{5/2}$ is $\frac{3}{10}$. Wait,the problem states $\frac{1}{10}(	an x)^A$. If the result is $\frac{3}{10}(	an x)^{5/2}$,there might be a typo in the question's coefficient. Assuming the form is $\frac{3}{10}(	an x)^A$,then $A = \frac{5}{2}$.

If $\int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x = \frac{3}{2}(\tan x)^B + \frac{1}{10}(\tan x)^A + c$,then $A =$

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