# Integration problem

1. Jul 29, 2007

### 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 - 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.

2. Jul 29, 2007

### Kurdt

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

3. Jul 29, 2007

### 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?

4. Jul 29, 2007

### Kurdt

Staff Emeritus
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.

5. Jul 29, 2007

### 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 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

Last edited: Jul 29, 2007
6. Jul 29, 2007

### Kurdt

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

7. Jul 29, 2007

### 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)) ?

8. Jul 29, 2007

### 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

9. Jul 29, 2007

### benedwards2020

Thankyou.... And many thanks for you help.