I=∫tanxsec2x dxtanx=t⇒sec2x.dx=dtI=∫tdt=23t32+c=23tan32x+c

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$\int \sqrt{\tan x} \sec^{2} x d x =$

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$\frac{2}{3} {(\cos x)}^{\frac{3}{2}} + C$

${(\tan x)}^{\frac{3}{2}} + C$

$\frac{2}{3} {(\tan x)}^{\frac{3}{2}} + C$

$\frac{1}{3} {(\tan x)}^{\frac{3}{2}} + C$

answer is D.

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Detailed Solution

$\begin{array}{l} I = \int \sqrt{\tan x} \sec^{2} x d x \\ \tan x = t \\ \Rightarrow \sec^{2} x . d x = d t \\ I = \int \sqrt{t} d t = \frac{2}{3} t^{\frac{3}{2}} + c = \frac{2}{3} \tan^{\frac{3}{2}} x + c \end{array}$

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