tan-3-x-sec-x-dx-

Question Number 125889 by bramlexs22 last updated on 15/Dec/20

\int \frac{\tan^{3} x}{\sqrt{\sec x}} d x ?

Answered by bobhans last updated on 15/Dec/20

\int \frac{\tan x (\sec^{2} x - 1)}{\sqrt{\sec x}} d x =

[\sec x = u^{2} \Rightarrow \tan x d x = \frac{2 d u}{u}]

I = \int \frac{(u^{4} - 1)}{u} (\frac{2 d u}{u}) = 2 \int (u^{2} - u^{- 2}) d u

= 2 (\frac{1}{3} u^{3} + \frac{1}{u}) + c

= \frac{2 \sqrt{\sec^{3} x}}{3} + \frac{2}{\sqrt{\sec x}} + c

Answered by Dwaipayan Shikari last updated on 15/Dec/20

\int \frac{{t a n x s e c}^{2} x}{\sqrt{s e c x}} - \frac{t a n x}{\sqrt{s e c x}} d x s e c x = t^{2} \Rightarrow s e c x t a n x = \frac{d t}{d x}

= 2 \int t^{2} d t - \int \frac{1}{t^{2}} = \frac{2}{3} t^{3} + \frac{2}{t} = \frac{2}{3} \sqrt{{s e c}^{3} x} + \frac{2}{\sqrt{s e c x}} + C