Improper Integral with trig integral

In summary, the improper integral ∫(dx/((1+x^2)^2) from 0 to ∞ converges and evaluates to ∏/4. The solution involves using the substitution x=tan(θ) and then using the trig identity cos^2(θ) = (1/2)(cos(2θ)+1) to solve the integral. The limits of integration should be from -∏/2 to ∏/2.
  • #1
shubox
3
0

Homework Statement


∫(dx/((1+x^2)^2) from 0 to ∞
Determine whether the improper integral converges and if so, evaluate it.

Homework Equations


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

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:
Physics news on Phys.org
  • #2
shubox said:

Homework Statement


∫(dx/((1+x^2)^2) from 0 to ∞
Determine whether the improper integral converges and if so, evaluate it.


Homework Equations


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

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

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
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
shubox said:
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

Not quite.

The denominator is (sec2(θ))2 .

Then use cos2(θ) = (1/2)(cos(2θ) + 1) .
 
  • #5
By the way, [itex]tan(\theta)[/itex] goes to infinity as [itex]\theta[/itex] goes to [itex]\pi/2[/itex]. So once you have made the substitution, instead of going to infinity, you limit should be going to [itex]\pi/2[/itex].
 
  • #6
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.
 

1. What is an improper integral with trigonometric integrals?

An improper integral with trigonometric integrals is a type of integral where the upper or lower limit of integration is infinite or the integrand contains trigonometric functions.

2. How is an improper integral with trigonometric integrals evaluated?

An improper integral with trigonometric integrals is evaluated using techniques such as substitution, integration by parts, or trigonometric identities.

3. What are the conditions for an improper integral with trigonometric integrals to converge?

For an improper integral with trigonometric integrals to converge, the integrand must approach zero as the limit of integration approaches infinity or negative infinity.

4. Can an improper integral with trigonometric integrals have a finite value?

Yes, an improper integral with trigonometric integrals can have a finite value if the integral converges.

5. What are some real-life applications of improper integrals with trigonometric integrals?

Improper integrals with trigonometric integrals are used in various fields of science and engineering, such as physics, astronomy, and signal processing, to model and solve problems involving periodic functions and infinite quantities.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
747
  • Calculus and Beyond Homework Help
Replies
3
Views
327
  • Calculus and Beyond Homework Help
Replies
2
Views
860
  • Calculus and Beyond Homework Help
Replies
7
Views
923
  • Calculus and Beyond Homework Help
Replies
10
Views
403
  • Calculus and Beyond Homework Help
Replies
2
Views
826
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
739
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
Back
Top