How do you find the indefinite integral of ${(tan x)}^{\frac{1}{2}} {(sec x)}^{2} d x$ $(tan x)^(1/2)(sec x)^2 dx$ ?

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Answer 1 · 2015-05-08T03:10:45

Use substitution.

Let $u = tan x$ $u=tanx$ then $d u = {sec}^{2} x d x$ $du=sec^2x dx$

So the integral becomes

$\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}} + C$ $int u^(1/2)du= 2/3u^(3/2)+C$

Undoing the substitution gets us

$\int {(tan x)}^{\frac{1}{2}} {sec}^{3} x d x = \frac{2}{3} {(tan x)}^{\frac{3}{2}} + C$ $int (tanx)^(1/2)sec^3x dx = 2/3 (tanx)^(3/2) + C$

Check the answer by differentiating.

How do you find the indefinite integral of (tanx)12(secx)2dx(tan x)^(1/2)(sec x)^2 dx?