Integrating product of trig functs with different args

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

 

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

{-integral sigh

{vdu=v*u -{dv*u

 

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