∫ se c ^ 2 x · cot(4)/(3)xdx is equal to

Given integral can be written as

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

Let $\tan x = t$

$⇒ \sec^{2}xdx = dt$

$⇒ I = ∫ {t}^{- \dfrac{4}{3}}dt$

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

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

$⇒ I = - \dfrac{3}{{t}^{\dfrac{1}{3}}} + C$

$⇒ I = - 3\tan^{- \dfrac{1}{3}}x + C .$