I=∫dxSin3x cos5xdx=∫dxSin xcos x3cos8x=∫sec4x dx(tan x)3=∫1+tan2xsec2x dx(tan x)3/2 put tan x=tsec2x dx=dtI=∫1+t2dtt3/2=∫t-3/2+t1/2dt=t-1/2-12+t3/232+C-2tanx+23(tan x)3/2+C=23(tan2x)-3tan x+C

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$\int \frac{d x}{\sqrt{\sin^{3} x \cos^{5} x}} =$

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$\frac{2}{3} (\frac{\tan^{2} x - 1}{\sqrt{\tan x}}) + c$

$\frac{2}{3} (\frac{\tan^{2} x - 2}{\sqrt{\tan x}}) + c$

$\frac{2}{3} (\frac{\tan^{2} x - 3}{\sqrt{\tan x}}) + c$

$\frac{2}{3} (\frac{\tan^{2} x - 4}{\sqrt{\tan x}}) + c$

answer is C.

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Detailed Solution

$\begin{matrix} I = \int \frac{dx}{\sqrt{S i n^{3} x \cos^{5} x}} dx \\ = \int \frac{dx}{\sqrt{{(\frac{Sin x}{\cos x})}^{3} \cos^{8} x}} \\ = \int \frac{\sec^{4} x dx}{\sqrt{{(\tan x)}^{3}}} \\ = \int \frac{(1 + \tan^{2} x) \sec^{2} x dx}{{(\tan x)}^{3 / 2}} \begin{matrix} put \tan x = t \\ \sec^{2} x dx = dt \end{matrix} \end{matrix} \begin{matrix} I = \int \frac{(1 + t^{2}) dt}{t^{3 / 2}} \\ = \int (t^{- 3 / 2} + t^{1 / 2}) dt \\ = \frac{t^{- 1 / 2}}{\frac{- 1}{2}} + \frac{t^{3 / 2}}{\frac{3}{2}} + C \\ \frac{- 2}{\sqrt{tanx}} + \frac{2}{3} {(\tan x)}^{3 / 2} + C \\ = \frac{2}{3} [\frac{(\tan^{2} x) - 3}{\sqrt{\tan x}}] + C \end{matrix}$