# Improper Integral with trig integral

1. Oct 23, 2011

### shubox

1. The problem statement, all variables and given/known data
∫(dx/((1+x^2)^2) from 0 to ∞
Determine whether the improper integral converges and if so, evaluate it.

2. Relevant equations
1+ tan^2(x) = sec^2(x)
1/sec(x) = cos(x)

3. The attempt at a solution
Initially I had no idea how to approach this problem. The answer in the back of the book is ∏/4, which tells me that maybe trig integrals are involved. So i started off with:
lim(R→∞) ∫(dx/((1+x^2)^2) from 0 to R.
x=tan(x)
lim(R→∞) ∫(dx/((1+tan^2(x))^2) from 0 to R.
=lim(R→∞) ∫(dx/(sec^4(x)) from 0 to R.
I do not know where to go from here. Any help would be appreciated

Last edited: Oct 23, 2011
2. Oct 23, 2011

### SammyS

Staff Emeritus
It's not a good idea to use the same variable in your substitution.

Let x=tan(θ) then dx = sec2(θ) dθ .

After finding the anti-derivative in terms of θ, change back to x, plug in the limits of integration, then take the limit.

3. Oct 23, 2011

### shubox

Oh I see now. Subsituting in dx cancels out sec2(θ) which then results in θ. Replacing θ with x gives arc tangent and then solving the integral from -∞ to ∞ results in ∏.
Thanks for helping

4. Oct 24, 2011

### SammyS

Staff Emeritus
Not quite.

The denominator is (sec2(θ))2 .

Then use cos2(θ) = (1/2)(cos(2θ) + 1) .

5. Oct 24, 2011

### HallsofIvy

Staff Emeritus
By the way, $tan(\theta)$ goes to infinity as $\theta$ goes to $\pi/2$. So once you have made the substitution, instead of going to infinity, you limit should be going to $\pi/2$.

6. Oct 25, 2011

### shubox

Ah I forgot the quantity squared. But after doing that I got the same thing: 1/2(cos(2θ)+1)
Taking the integral of this from ∏/2 to -∏/2 results in ∏/2 which is the answer in the back of the book.

sweeet, makes sense now. thanks alot.