Prove Integral of 1/(t^2+a^2) = 1/a tan-1(t/a) + c

[Sidthewall]

Homework Statement

How do i prove the integral of 1/(t^2+a^2) = 1/a tan-1(t/a) + c

Homework Equations

The Attempt at a Solution

 

[Sidthewall]

oh yeah and sub in au as t

 

[Mark44]

Homework Statement

How do i prove the integral of 1/(t^2+a^2) = 1/a tan-1(t/a) + c

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

Homework Equations

The Attempt at a Solution

What have you tried? What techniques do you know?

oh yeah and sub in au as t

What are you asking here?

 

[Sidthewall]

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

 

[hunt_mat]

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

Mat

 

[Sidthewall]

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

 

[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

 

[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?

 

[hunt_mat]

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

Mat

 

[Mark44]

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.

 

[Sidthewall]

matt can u show me step by step

 

[Sidthewall]

yeah i do

 

[Sidthewall]

integrate that then since u=t/a sub that back in... i don't get the algebra part though

 

[Mark44]

integrate that then since u=t/a sub that back in... i don't get the algebra part though

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.

 

[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
$$\frac{1}{a}\tan^{-1}\frac{x}{a}$$

 

[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

 

[Mark44]

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

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

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

 

[vela]

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?

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

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

as being equal to arctan u. If you start with

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

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

 

[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

 

[Mark44]

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

 

[Char. Limit]

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

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

as being equal to arctan u. If you start with

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

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

So then, while obscure and not really useful, this is a valid definition for the arctangent function?

 

[vela]

It's a valid expression for arctan and could be useful when working with complex numbers.

 

[HallsofIvy]

You also have an "a" in the numerator: if t= au, then dt= a du.