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$\int \sec^{2 / 3} x {cosec}^{4 / 3} x d x =$

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$- 3 (\tan x)^{1 / 3} + c$

$- 3 (\tan x)^{- 1 / 3} + c$

$3 (\tan x)^{- 1 / 3} + c$

$(\tan x)^{- 1 / 3} + c$

answer is B.

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Detailed Solution

$\int \sec^{2 / 3} x {cosec}^{4 / 3} x d x$

$= \int \frac{d x}{\sin^{4 / 3} x \cos^{2 / 3} x}$

Multiplying and Dividing by $\cos^{4 / 3} x$ to Dinominator we get

$\begin{array}{l} = \int \frac{\sec^{2} x d x}{\tan^{4 / 3} x} = \int \frac{d t}{t^{4 / 3}} = \frac{t^{- 1 / 3}}{(- 1 / 3)} + c \\ {Putting \tan x = t \Rightarrow \sec^{2} x d x = d t\} \\ = - 3 (\tan x)^{- 1 / 3} + c . \end{array}$

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