Modifying integral so that it matches the table

[Saladsamurai]

Homework Statement

I am trying to integrate:

\int e^x\sin(T+x)\,dx

where T is independent of x. I have the following formula from a table of integrals:

I am a little confused as to how I can use this? The arguments of my functions are not identical so I am not sure how to go about this? Can I actually use this formula?

 

[fzero]

Just use

\sin y = \frac{e^{iy}-e^{-iy}}{2i}.

You also could use the trig identity for sin of a sum to use the formulae from integral tables.

 

[Saladsamurai]

Just use

\sin y = \frac{e^{iy}-e^{-iy}}{2i}.

You also could use the trig identity for sin of a sum to use the formulae from integral tables.

hi fzero

For your latter suggestion: Which trig identity? sine of sum goes to a product of sine and cosine, so my formula would still not apply ...

 

[Mark44]

I would use an ordinary substitution for starters: u = x + T, du = dx. The integral then becomes:
e^{-T}\int e^u\sin(u)\,du

That one can be done using two integration by parts.

 

[fzero]

hi fzero

For your latter suggestion: Which trig identity? sine of sum goes to a product of sine and cosine, so my formula would still not apply ...

Yes, but I'd assume there's an analogous formula for the cos integral. You can also use a change of variables u=x+T, for which e^x = e^{-T} e^u.

 

[Saladsamurai]

I would use an ordinary substitution for starters: u = x + T, du = dx. The integral then becomes:
e^{-T}\int e^u\sin(u)\,du

That one can be done using two integration by parts.

Yes, but I'd assume there's an analogous formula for the cos integral. You can also use a change of variables u=x+T, for which e^x = e^{-T} e^u.

Ooooo you guys are sneaky!

 

[SammyS]

Just use

\sin y = \frac{e^{iy}-e^{-iy}}{2i}.

You also could use the trig identity for sin of a sum to use the formulae from integral tables.

This won't help with using the table of integrals.


But, yes, you can use this formula from the table of integrals.

Use the substitution: u=T+x\quad\to\quad du=dx\ ,\quad\text{also gives: }x=u-T

Therefore: e^x=e^{u-T}=e^{-T}\,e^u\,. and e^-T is just a constant coefficient.