How to calculate this? `int_0^(2pi)sin(mx)cos(nx)dx`; `m,n in ZZ`.

Using

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

so

`1/2(sin(m+n)x+sin(m-n)x)=sinmx cos nx`

now

`int_0^(2pi)sinmx cos nx dx = 1/2int_0^(2pi)(sin(m+n)x+sin(m-n)x)dx =`

`=-[Cos((m - n) x)/(2 (m - n)) +Cos((m + n) x)/(2 (m + n))]_0^(2pi)=`

`=(sin^2(m-n)pi)/(m-n)+(sin^2(m+n)pi)/(m+n) = 0`

Note that

`lim_(xi->0)(sin^2 xi pi)/xi = 0`

NOTE:

`[-Cos(2(m + n) x)/(2 (m + n))]_0^(2pi) = -Cos((m + n) pi)^2/(2 (m + n)) + Sin((m + n)pi)^2/(2 (m + n))+1/(2(m+n)) =Sin((m + n) pi)^2/(m + n)`

`int_0^(2pi) f(x)dx = int_(-pi)^pi f(x+pi)dx`

but here `f(x+pi)=(-1)^(m+n)f(x)` and also

`f(x)=-f(-x)` is an odd function so

`int_0^(2pi) f(x)dx = (-1)^(n+m)int_(-pi)^pi f(x) dx = 0`

Substitute `t=x-pi` so `x = t-pi` and `dx = dt`.

The limits of integration become `-pi` and `pi` and the new integrand is an odd function.

`sin(mx) = sin(mt-mpi) = sin(mt)cos(mpi)` which is an odd function of `t`

`cos(nx) = cos(nt)cos(npi)` which is an even function of `t`.

`int_0^(2pi)sin(mx)cos(nx)dx = int_-pi^pi sin(mt)cos(mpi)cos(nt)cos(npi) dt`

The integrand on the right is odd, so the integral from `-a` to `a` is `0`.