# Integration problem

I need to find the indefinite integral

$$\int \frac{sec^2 (x)}{\sqrt{1-tan^2 (x)}}$$

Now, I'm not sure which method to use here.... I think that the quotient and the square root is confusing me here. I can certainly integrate the numerator - thats not the problem, I'm not sure how to handle the demoninator. Can anyone give me any pointers on how to tackle this? It's the last question on the paper and I'd like to attempt to answer it.

Kurdt
Staff Emeritus
Gold Member
How about substituting $u=tan(x)$

ok, so this will give me

$$\frac{sec^2 (x)}{\sqrt{1-u^2}}$$

where

$$\frac{du}{dx}=sec^2 dx$$

And to get rid of the root, will give us

$$\frac{sec^2 (x)}{(1-u^2)^\frac{1}{2}}$$

Am I right so far?

Kurdt
Staff Emeritus
Gold Member
Yeah almost there. What you end up with after the substitution is this integral.

$$\int \frac{du}{(1-u^2)^\frac{1}{2}}$$

To deal with this will require another substitution and it will turn the integral into something trivial.

HINT: think trig identities.

Ok, that makes sense... Now I have a list of trig identities and I can see that

$$arctan(x) = \frac{1}{1+x^2}$$

Again its that half power thats confusing me....

Unless arcsin(x) is correct. Hmm

$$arcsin(x) = \frac{1}{\sqrt{1-x^2}}$$

which is the same as

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

Which is very close to what we have

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Kurdt
Staff Emeritus
Gold Member
That will link in later but if for now you try the substitution $u=sin(t)$ and see where that takes you.

Hmm... I don't think I'm following you for the second substitution.... My books don't give me those examples.

If i say that

$$\frac{1}{(1-u^2)^\frac{1}{2}} = arcsin$$

But that u = tan(x)

Surely we can conclude that the answer is

arcsin(tan(x)) ?

Kurdt
Staff Emeritus
That is the answer. If you do the sin substitution you can confirm this for yourself which is what I was aiming at. But hey well done for spotting that 