∫ s e c 2 x · cot 4 3 x d x is equal to

Given integral can be written as

$I=\int \frac{{\sec }^{2}x}{{\left(\tan x\right)}^{\frac{4}{3}}} dx$

Let $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow I=\int t^{-\frac{4}{3}} dt$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,$ we get

$I=\frac{t^{-\frac{1}{3}}}{\left(-{\displaystyle \frac{1}{3}}\right)}+ C$

  $\Rightarrow I=-\frac{3}{t^{{\displaystyle \frac{1}{3}}}}+ C$

$\Rightarrow I=-3{\tan }^{-\frac{1}{3}}x+C.$