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

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Answer 1 · 2015-05-08T03:10:45

Use substitution.

Let #u=tanx# then #du=sec^2x dx#

So the integral becomes

#int u^(1/2)du= 2/3u^(3/2)+C#

Undoing the substitution gets us

#int (tanx)^(1/2)sec^3x dx = 2/3 (tanx)^(3/2) + C#

Check the answer by differentiating.