# Integration problem

benedwards2020
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 - that's 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.

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

benedwards2020
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?

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.

benedwards2020
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 that's 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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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.

benedwards2020
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)) ?

Staff Emeritus