# Integrating product of trig functs with different args

1. Feb 17, 2007

### electric.avenue

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution

2. Feb 17, 2007

### 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

3. Feb 17, 2007

### 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

4. Feb 17, 2007

### 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.

5. Sep 5, 2007

### electric.avenue

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