Integration with Trig Substitution

In summary, bob suggests that he should start on a calculus course before attempting intro physics, and that he is still a few months away from finishing all the courses.
  • #1
Oerg
352
0

Homework Statement


Ok, so I was doing a problem on the electric field strength of a continuous charge distribution and I arrived at this seemingly easy integral

[tex] \int \frac{1}{({l^2+a^2})^\frac{3}{2}} dl[/tex]
sorry the latex is lagging badly, you can see the correct integral by clicking on it. it is 1/l squared + a squared with the denominator to the power of 3/2.

Homework Equations


The Attempt at a Solution


I do not know how to solve this... at all, and I am pretty sure that this integral is correct.
 
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  • #2
Recheck your latex and include which variable you're integrating wrt.
 
  • #3
This requires nothing more than a trig substitution.

[tex]l=a\tan\theta[/tex]
[tex]dl=a\sec^{2}\theta d\theta[/tex]

[tex]*\cos^2 x=\frac{1}{2}(1+\cos{2x})[/tex]
 
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  • #4
omggggg... this is killing me! wth invented integration

The final answer was supposed to be trigo free. Can you hint a little more, I am a little confused =x

btw it is supposed to be a definite integral from infinity to negative infinity.

oh yes, I am able to convert [tex]cos\theta[/tex] and [tex]sin\theta[/tex] to [tex]\frac{a}{(l^2+a^2)^\frac{1}{2}}[/tex] or[tex] \frac{l}{(l^2+a^2)^\frac{1}{2}}[/tex]PROVIDED

that [tex]\theta[/tex] refers to the angle that [tex](l^2+a^2)^{0.5}[/tex] makes to the vertical.

By the way, this is a question of fidning the electric field strength of an infinitely long charged wire.
 
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  • #5
You don't have to use trig sub., but I find it the easiest. Here is another thread with the exact problem you are doing except with different letters.

bob suggests to factor out the constant ... if you need more help post

https://www.physicsforums.com/showthread.php?t=220687
 
  • #6
SIGHHHHHHHH

I guess I should start on my calculus 1 course first before attempting intro physics.Something i take comfort in though, I managed to come up with the electric field strength after I worked with the correct solution in the page your provided.
 
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  • #7
Oerg said:
SIGHHHHHHHH

I guess I should start on my calculus 1 course first before attempting intro physics.
You haven't taken Calculus? Does this Intro course only require like a wimpy Intro or Elements to Calculus course?
 
  • #8
nope, I am actualyl studying for my exemption test where I will get exempted rfom first course modules if I pass the tests.

I did study some calculus in high school which was required from the syllabus, either I am rusty, or they did not cover enough ground.
 
  • #9
Oerg said:
nope, I'm actually studying for my exemption test where I will get exempted from first course modules if I pass the tests.

I did study some calculus in high school which was required from the syllabus, either I am rusty, or they did not cover enough ground.
What courses are in the "first course modules"?

Referring to Calculus: Trust me, it's worth taking the course. If you need more convincing, post in the Academic sub-forum. What you learn in a College Calculus course is far more in-depth than what you learned in HS.
 
  • #10
six modules in total:

Cal 1, linear algebra, diff equations

mechanics, Waves, EM
 
  • #11
Oerg said:
six modules in total:

Cal 1, linear algebra, diff equations

mechanics, Waves, EM
... You want to skip all those courses?
 
  • #12
If I could... Why not? I still have 5 months left,

Im done with mechanics which was easy, halfway through EM, guess i should start on Cal 1 soon.

oh yes, diff equations, I guess I will need them badly when doing electric potentials.
 

FAQ: Integration with Trig Substitution

What is integration with trig substitution?

Integration with trig substitution is a technique used in calculus to solve integrals that involve functions with trigonometric expressions. It involves substituting a trigonometric function for a variable in the integral, simplifying the expression, and then using trigonometric identities to solve the integral.

When should I use trig substitution for integration?

Trig substitution is typically used when the integral involves a radical expression, or when the integral involves a quadratic expression with a square root. It is also used when the integral involves a product of a trigonometric function and a polynomial.

What are the steps for integration with trig substitution?

The steps for integration with trig substitution are as follows:1. Identify the appropriate substitution by looking at the expression inside the integral and determining which trigonometric function can be substituted for a variable.2. Substitute the trigonometric function for the variable and rewrite the integral in terms of the new variable.3. Simplify the expression using trigonometric identities.4. Solve the new integral using standard integration techniques.5. Substitute the original variable back into the solution to obtain the final answer.

What are the common trigonometric identities used in integration with trig substitution?

Some common trigonometric identities used in integration with trig substitution include:1. Pythagorean identities: sin²x + cos²x = 1 and tan²x + 1 = sec²x2. Double angle identities: sin2x = 2sinx*cosx and cos2x = cos²x - sin²x3. Half angle identities: sin²(x/2) = (1 - cosx)/2 and cos²(x/2) = (1 + cosx)/2

Are there any tips for solving integrals with trig substitution?

Yes, here are some tips that can help when solving integrals with trig substitution:1. Always check if the integral can be solved using other techniques, such as integration by parts, before resorting to trig substitution.2. Pay attention to the limits of integration and make sure to adjust them accordingly when substituting for a new variable.3. Keep a list of common trigonometric identities handy to simplify the expression.4. Practice, practice, practice! With enough practice, you will become more familiar with which substitution to use for different types of integrals.

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