How can I integrate this one?

Question

anonymous · Answer

$$\int\limits_{}^{} \frac{ \sec ^{2}xdx }{ \sqrt{5-\sec ^{2}x} } $$

john_es · Answer

You can choose a suitable change of vairable. For example, 
$$\tan x=t\Rightarrow \sec^2xdx=dt$$
Also, you find 
$$sec^2x=1+t^2$$
Then, making the replacements in the integral,
$$\int\frac{\sec^2x\ dx}{\sqrt{5-\sec^2x}}=\int\frac{dt}{\sqrt{4-t^2}}=\arcsin(t/2)$$
Replacing t
$$t=\arctan x\Rightarrow \int\frac{\sec^2x\ dx}{\sqrt{5-\sec^2x}}=\arcsin\left(\frac{1}{2}\arctan x\right)$$
You should add the usual constant K for the indefinite integrals.

anonymous · Answer

I'm still trying to understand the step by step procedures especially on choosing the replacements. But still, thank you very much!

anonymous · Answer

The value of $$u ^{2}= \tan ^{2}x$$ then the value of u=tanx.
Why did it became an arctan on the final answer?

john_es · Answer

It was my fault, 
$$t=\tan x$$
So, the solution should be,
$$\arcsin\left(\tan x/2\right)$$
Sorry for the trouble.

anonymous · Answer

No, it is alright (without your help I wouldn't be able to answer it). Thanks for the help!