How do you tackle this integral

  • Thread starter Gauss M.D.
  • Start date
  • Tags
    Integral
In summary: Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:##y = \int {dx \over (1 - x^2)^2}####x =? \; sin\theta####dx = cos\theta \; d\theta####y = \int sec^3\theta \; d\theta#### = {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C#### = {1 \over 2} [sec\theta \; tan\theta + l
  • #1
Gauss M.D.
153
1

Homework Statement



Find F(x)

f(x) = 1/(1+x2)2)

Homework Equations





The Attempt at a Solution



It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?
 
Physics news on Phys.org
  • #2
Gauss M.D. said:

Homework Statement



Find F(x)

f(x) = 1/(1+x2)2)

Homework Equations


The Attempt at a Solution



It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?

Perfect candidate for a trig sub. Use ##x = tan(θ)## so that ##dx = sec^2(θ)dθ##.
 
  • Like
Likes 1 person
  • #3
well, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?
 
  • #4
Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
 
Last edited:
  • #5
Gauss M.D. said:
Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?

its a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?
 
  • #6
No, I was referring to Zondrinas trig sub suggestion!
 
  • #7
Gauss M.D. said:
Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?

I don't see any discontinuity in the integrand at all, so I don't think it would matter.
 
  • #8
Right right, we're using tan and not some other trig function. My bad again. Thanks a ton :)
 
  • #9
Gauss M.D. said:
Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?

That is a very good question, I don't know the answer to it. I'm going to investigate this.
 
  • #10
verty said:
That is a very good question, I don't know the answer to it. I'm going to investigate this.

Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:

##y = \int {dx \over (1 - x^2)^2}##

##x =? \; sin\theta##
##dx = cos\theta \; d\theta##

##y = \int sec^3\theta \; d\theta##
## = {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C##
## = {1 \over 2} [sec\theta \; tan\theta + ln| sec\theta + tan\theta | ] + C##
## = {1\over 2}{x \over 1 - x^2} + {1\over 4} ln | {1+x\over 1-x}| + C##On the other hand, one can make a translation:

##x = u-1##
##dx = du##

##y = \int {dx \over (1 - x^2)^2} = \int{ dx \over (1+x)^2 (1-x)^2} = \int {du \over u^2(2-u)^2}##

After some partial fraction magic:

##y = \int {1 \over 4(2-u)^2} + {1 \over 4u^2} + {1 \over 4u} + {1 \over 4(2-u)} du##
## = {1 \over 4} [ {1 \over 2-u} - {1\over u} + ln|u| - ln|2-u| ] + C##
## = {1 \over 4} [ {2(u-1) \over u(2-u)} + ln| {u \over 2-u} | ] + C##
## = {1 \over 2} {x \over 1 - x^2} + {1 \over 4} ln| {1+x \over 1-x} | + C##

I can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.
 

1. How do you approach solving an integral?

When tackling an integral, it is important to first identify the type of integral you are dealing with. This could be a definite or indefinite integral, as well as a trigonometric, exponential, or polynomial integral. Once you have determined the type of integral, you can then choose an appropriate method for solving it, such as substitution, integration by parts, or partial fractions.

2. What steps do you follow to solve an integral?

The general steps for solving an integral are as follows: 1) Identify the type of integral, 2) Simplify or manipulate the integrand if necessary, 3) Apply an appropriate method for solving the integral, 4) Check your answer by taking the derivative, and 5) Add any necessary constants for indefinite integrals.

3. How do you decide which method to use when solving an integral?

The choice of method for solving an integral depends on the type of integral and the complexity of the integrand. For example, substitution is often used for integrals involving trigonometric functions, while integration by parts is useful for integrals involving products of functions. It may also be helpful to try different methods and see which one gives the simplest result.

4. What are some common mistakes to avoid when solving integrals?

Some common mistakes to avoid when solving integrals include forgetting to add the constant of integration, making algebraic errors, and using the wrong method for the type of integral. It is also important to be careful with signs and to double check your work for accuracy.

5. How can I improve my skills in solving integrals?

The best way to improve your skills in solving integrals is through practice. Start with simple integrals and gradually work your way up to more complex ones. It can also be helpful to study and understand the underlying concepts and techniques used in solving integrals, and to seek help from a tutor or teacher if needed.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
853
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
44
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
Back
Top