1 Answer

Xolox · Answer 1 · 2021-10-09T10:25:03+0000

Answer:

$2(\tan(x))^(1)/(2)+C$

Step-by-step explanation:

Let
$u=\tan(x)$ .

Therefore,
$du=\sec^2(x)dx$ or
$du=(\sec(x))^2dx$ .

$\int ((\sec(x))^2 dx)/(√(\tan(x)))$

Putting the integral in terms of
now:

$\int (du)/(√(u))$

or

$\int (1)/(√(u)) du$

or

$\int (1)/(u^(1)/(2))du$

or

$\int u^(-1)/(2) du$

Let's integrate now:

$\frac{u^{(-1)/(2)+1}}{(-1)/(2)+1}+C$

$\frac{u^{(1)/(2)}}{(1)/(2)}+C$

$(2)/(2) \frac{u^{(1)/(2)}}{(1)/(2)}+C$

$\frac{2u^{(1)/(2)}}{1}+C$

$2u^{(1)/(2)}+C$

Putting back in terms of
:

$2(\tan(x))^(1)/(2)+C$

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