# Proving an integral

1. Jul 15, 2010

### Sidthewall

1. The problem statement, all variables and given/known data
How do i prove the integral of 1/(t^2+a^2) = 1/a tan-1(t/a) + c

2. Relevant equations

3. The attempt at a solution

2. Jul 15, 2010

### Sidthewall

oh yeah and sub in au as t

3. Jul 15, 2010

### Staff: Mentor

Here is the equation in LaTeX. Click anywhere in the equation to see what I did.
$$\int \frac{dt}{t^2 + a^2} = \frac{1}{a}tan^{-1}(t/a) + C$$
What have you tried? What techniques do you know?

4. Jul 15, 2010

### Sidthewall

What it is asking is to prove that integral by substituting t= au in the left hand side

5. Jul 15, 2010

### hunt_mat

Try the substitution t=a*tan u. Use the identity 1+tan^2u=sec^u.

Mat

6. Jul 15, 2010

### Sidthewall

i thought of that but the question said to use t=au

7. Jul 15, 2010

### hunt_mat

That only gets you to:
$$\frac{1}{a}\int\frac{du}{1+u^{2}}$$
What do you do then? My idea gets over that in one fell swoop.

Mat

8. Jul 15, 2010

### Char. Limit

This question made me think of a similar issue... but it's probably a very bad idea. I'm just checking to see how bad it is.

Could you make a definition of the inverse tangent function by factoring 1+u^2 into (1-iu)(1+iu), applying partial fractions, and integrating? Or this is a Bad Idea?

9. Jul 15, 2010

### hunt_mat

Very, you have turned a real integral into a complex one.

Mat

10. Jul 15, 2010

### Staff: Mentor

Sidthewall,
Do you know this integral formula?
$$\int \frac{dt}{t^2 + 1} =tan^{-1}(t) + C$$
The suggested substitution makes me think that this is an integral formula that you already know, so applying the substitution is not much more than a problem in algebra.

The substitution suggested by hunt_mat in post #5 is a more general approach that doesn't require that you know the formula above.

11. Jul 15, 2010

### Sidthewall

matt can u show me step by step

12. Jul 15, 2010

### Sidthewall

yeah i do

13. Jul 15, 2010

### Sidthewall

integrate that then since u=t/a sub that back in... i dont get the algebra part though

14. Jul 15, 2010

### Staff: Mentor

So far, you haven't shown us anything that you have done. We are not here to do your work for you, but to help you do it. Show us what you have done and where you're stuck.

15. Jul 15, 2010

### hunt_mat

To integrate:
$$\int\frac{dx}{x^{2}+a^{2}}$$
Use $$x=a\tan u$$, then $$dx=a\sec^{2}udu$$. But
$$x^{2}+a^{2}=a^{2}\tan^{2}u+a^{2}=a^{2}(1+\tan^{2}u)=a^{2}\sec^{2}u$$
The integral becomes:
$$\frac{1}{a}\int\frac{\sec^{2}u}{\sec^{2}u}du=\frac{1}{a}\int dx=\frac{u}{a}+C$$
However $$u=\tan^{-1}(x/a)$$ and the solution is
$$\frac{1}{a}\tan^{-1}\frac{x}{a}$$

16. Jul 15, 2010

### Sidthewall

k what i am stuck on is the algebra

so this is what i have

1/((au)^2 + a^2)
= 1/(( (a)^2)(u^2) + a^2)

I factored out a^2 to get u^2 + 1,,,, but the constant in the integral is suppose to be 1/a... that is were I am stuck

17. Jul 15, 2010

### Staff: Mentor

$$\frac{1}{(au)^2 + a^2} = \frac{1}{a^2u^2 + a^2} = \frac{1}{a^2(u^2 + 1)}$$

Since u = t/a, then du = dt/a. Is this what you're missing?

18. Jul 15, 2010

### vela

Staff Emeritus
It'll work, though you may not recognize the answer

$$\frac{1}{2i}\log\left(\frac{1+iu}{1-iu}\right)$$

$$\tan x = \frac{\sin x}{\cos x}=\frac{(e^{ix}-e^{-ix})/(2i)}{(e^{ix}+e^{-ix})/2} = u$$

and solve for x, you'll end up with the same expression.

19. Jul 15, 2010

### Sidthewall

that's what i was missin u = t/a, then du = dt/a
and how do u get ur math statements in proper form so it's easier on the eyes

20. Jul 15, 2010

### Staff: Mentor

Several of us are using LaTeX in this thread, which produces output like this:
$$\frac{1}{(au)^2 + a^2} = \frac{1}{a^2u^2 + a^2} = \frac{1}{a^2(u^2 + 1)}$$

Click anywhere in the equation and a window will pop open showing the LaTeX script. There's more info here: https://www.physicsforums.com/showthread.php?t=386951 [Broken]

Last edited by a moderator: May 4, 2017