Solving Indefinite Integral: $\int \frac{sec^2 x}{\sqrt{1-tan^2 x}}dx$

[benedwards2020]

Homework Statement

Find the indefinite integral of

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


I'm stuck on how to proceed with the denominator. I know that

$$\sqrt{1+tan^2 x}$$

is equivalent to $$\\sec\theta$$

but I can't seen to find an equivalent to what I have. Can anyone give me any pointers?

 

[Dick]

What's the derivative of arcsin?

 

[George Jones]

Can you make a sustitution involving $u$ such that $du = sec^2 x dx$?

 

[benedwards2020]

The derivative of arcsin would be

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

 

[Dick]

Right. Suppose we put tan(x) in place of x?

 

[benedwards2020]

This would give us

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

Which leaves us with a term equivalent to us getting rid of the square root

 

[Dick]

d/dx(arcsin(tan(x)))=?. Don't forget the chain rule.

 

[benedwards2020]

Hmm... This is where I'm losing it I think. The chain rule will give us

$$f(x)=arcsin$$
$$f'(x)=\frac{1}{(1-x^2)^\frac{1}{2}}$$
$$g(x)=tan(x)$$
$$g'(x)=1+tan^2 (x)$$


$$\frac{d}{dx}arcsin(tan(x))=\frac{1}{(1-x^2)^\frac{1}{2}} \times tan(x) \times \left( 1+tan^2 (x) \right)$$

Not sure if this is right but I've provided my values so you can see where I've gone wrong

 

[Dick]

I'm sorry. I think I'm going way too fast and confusing you. George's step by step suggestion is probably much better. The chain rule is (f(g(x)))'=f'(g(x))*g'(x). You have written f'(x)*g(x)*g'(x). Do you see the difference? And there is a more compact way to write 1+tan(x)^2. When you finish this go back and look at the route George would have led you on.

 

[benedwards2020]

Ok, is there a quick way of identifying what needs to be substituted. At the moment I'm doing it by trial and error, and obviously in an exam, that's not going to be very economical

 

[Dick]

You want something whose derivative is sec^2(x). Memorizing at least small list of derivative and integral forms is necessary to do this kind of stuff with any speed. You've already said that the derivative of tan(x) is tan^2(x)+1. This is certainly correct, but most people would probably respond with a different (but equivalent) answer. What does the wikipedia table of derivatives say?

 

[benedwards2020]

Just had a look at wikipedia. It shows that the more common answer for the derivative of tan(x) is

$$sec^2 (x)$$

or

$$\frac{1}{cos^2 (x)}$$

 

[Dick]

Exactly, so if u=tan(x) then du=sec^2(x)*dx.

 

[benedwards2020]

Ok, I'm happy with how we got to this point.

So what I have now is the integral of 'something' times sec^2(x) dx

Is this right? And now I need to find what makes up the equivalent of the original integral... Am I on the right lines here?

I think half the problem I'm having with this question is that my books don't give such detailed examples. I need to follow a lot more examples... Do you know of any good sites (apart from this one) that gives plenty of worked examples?

 

[Dick]

The point is that u=tan(x) might be a good substitution to do on the original integral.