How do you integrate $\int \sqrt{tan x} {sec}^{2} x d x$ $int sqrttanxsec^2xdx$ ?

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Answer 1 · 2016-11-29T19:45:27

Let $u = tan (x)$ $u = tan(x)$ , then $d u = {sec}^{2} (x) d x$ $du = sec^2(x)dx$ . Please see the explanation.

Let $u = tan (x)$ $u = tan(x)$ , then $d u = {sec}^{2} (x) d x$ $du = sec^2(x)dx$

$\int \sqrt{tan (t)} {sec}^{2} d x = \int (u^{\frac{1}{2}}) d u$ $intsqrt(tan(t))sec^2dx = int(u^(1/2))du$

Integrate:

$\int \sqrt{tan (t)} {sec}^{2} d x = \frac{2}{3} u^{\frac{3}{2}} + C$ $intsqrt(tan(t))sec^2dx = 2/3u^(3/2) + C$

Reverse the substitution:

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

How do you integrate int sqrttanxsec^2xdx∫√tanxsec2xdxint sqrttanxsec^2xdx?