Integrating product of trig functs with different args

[electric.avenue]

Homework Statement

How do you integrate a product of two trigonometric functions of x, when the argument is different, ie:

what is the integral of

sin x cos 3x


I ought to know this, but don't seem to be able to do it.

Thanks in adv!

Homework Equations

The Attempt at a Solution

 

[AlephZero]

Use the formulas for sin(a+b) and sin(a-b):

sin(a+b) = sin a cos b + cos a sin b
sin(a-b) = sin a cos b - cos a sin b

so

sin(a+b) + sin(a-b) = 2 sin a cos b

 

[transgalactic]

you can solve this by the formula
choose one part as v
choose the other as du

{-integral sigh

{vdu=v*u -{dv*u

 

[arildno]

Note that transgalactic's proposal will enable you to find the answer by entering an integrand loop of finite length.

To see what I mean with an "integrand loop", I'll take an example:

We are to compute:
I=\int_{0}^{1}e^{x}\cos(x)dx
we use partial integration to write this as:
I=\int_{0}^{1}e^{x}\cos(x)dx=e^{x}\cos(x)\mid^{x=1}_{x=0}+\int_{0}^{1}e^{x}\sin(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-\int_{0}^{1}e^{x}\cos(x)dx=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I
Therefore, we have gained the equation for I:
I=(e^{x}\cos(x)+e^{x}\sin(x))\mid_{x=0}^{x=1}-I
which is easily solved for I.

 

[electric.avenue]

Thanks for all the help and advice, everyone. I got an A in maths, btw.