Proving Integrals: e^(au)sin(bu), e^(au)cos(bu), sec^3u du

In summary, the conversation includes a request for proofs of three integrals: integral ( e^(au)sin(bu)du ), integral(e^(au)cos(bu)du), and integral(sec^3u du). The first two can be solved using integration by parts twice, and the third one can also be solved using integration by parts. The conversation also mentions trigonometric derivatives and suggests using them to solve the integrals. Lastly, there is a request for a link to proofs of integration by parts.
  • #1
nameVoid
241
0
looking for proofs of the following integrals

integral ( e^(au)sin(bu)du ) and integral(e^(au)cos(bu)du) and integral (sec^3u du)
 
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  • #2
The first two can be done by integrating by parts twice. After the second integration, you end up with the integral you started with, and you can solve algebraically for the integral. In both problems, start with u = e^(au) and dv = sin(bu)du or cos(bu)du. After integrating the first time, do it again, with u = e^(au) again, and dv being whatever is left.

The third one can also be done by integration by parts, I believe. Try u = sec u and dv = sec^2(u) du.
 
  • #3
also remember your trig derivatives:

d/dx(secx)=tanxsecx and
d/dx(tanx)=sec2x
 
  • #4
[tex]\int \exp(p x) dx = \frac{1}{p}\exp(px) + c[/tex]

Take p = a + b i and equate the real and imaginary parts of both sides. You should be able do this mentally in your head (multiplying the numerator and denominator by the complex conjugate of p isn't that difficult).
 
  • #5
im looking for somthing along the lines of integration by parts a link to proofs would be nice
 
  • #6
nameVoid said:
im looking for somthing along the lines of integration by parts a link to proofs would be nice
Your problem isn't a proof. All you need to do is carry out the integration, and I have given you a start on how to do that.
 
  • #7
Do you mean that you want a proof of the actual method of integration by parts? If so, then consider the product rule of differentiation:

d/dx(f*g)=f'*g + f*g'

Now integrate both sides of the equation and you get you proof.

Is that what you were looking for?
 

FAQ: Proving Integrals: e^(au)sin(bu), e^(au)cos(bu), sec^3u du

What is the process for proving integrals involving e^(au)sin(bu)?

The process for proving integrals involving e^(au)sin(bu) is to use integration by parts. This involves breaking the integral into two parts, one with e^(au) as the function and the other with sin(bu) as the function. Then, use the formula u dv = uv - ∫v du to find the integral of each part. The two resulting integrals can then be combined to solve for the original integral.

Can the same process be used for proving integrals involving e^(au)cos(bu)?

Yes, the same process of integration by parts can be used for proving integrals involving e^(au)cos(bu). Just like with e^(au)sin(bu), the integral can be split into two parts and then solved using the u dv formula. The resulting integrals can then be combined to find the original integral.

What about proving integrals involving sec^3u du? Is the process different?

Yes, the process for proving integrals involving sec^3u du is different. This type of integral requires the use of trigonometric identities and substitution. The first step is to use the identity sec^2u = 1 + tan^2u to rewrite the integral. Then, a substitution can be made, such as u = tan(x), to simplify the integral. Finally, the integral can be solved using the u-substitution rule.

Are there any special cases to consider when proving integrals involving e^(au)sin(bu) or e^(au)cos(bu)?

Yes, there are two special cases to consider. The first is when the value of b is equal to 0. In this case, the integral can be solved directly without using integration by parts. The second is when the value of b is equal to a. In this case, the integral can be solved using a substitution, such as u = au, to simplify the integral.

Is there a way to check if the solution to a proven integral is correct?

Yes, there is a way to check the solution to a proven integral. This can be done by differentiating the solution and comparing it to the original integrand. If the two are equal, then the solution is correct. Additionally, the solution can be graphed to visually confirm that it matches the original function.

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