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Q.

If $\int \sec^{2 / 3} x \cos e c^{4 / 3} x d x = m \tan^{- 1 / 3} x + C$ (where C is constant of integration) then $m =$

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a

$- \frac{1}{3}$

b

$- 3$

c

$\frac{4}{3}$

d

3

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{d x}{\cos^{\frac{2}{3} x} (\frac{\sin^{\frac{4}{3} x}}{\cos^{\frac{4}{3} x}}) \times \cos^{\frac{4}{3} x}} \\ = \int \frac{\sec^{2} x d x}{\tan^{\frac{4}{3}} x} \end{array}$

$\begin{array}{l} \tan x = z \Rightarrow \sec^{2} x d x = d z \\ = \int z^{- \frac{4}{3}} d z = \frac{z^{- \frac{1}{3}}}{- \frac{1}{3}} + C \\ = - 3 {(\tan x)}^{\frac{1}{- 3}} + C \end{array}$

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