Question

How do you integrate `inte^(cos)(t) (sin 2t) dt` between `a = 0`, `b = pi`?

Answer 1

This integral has to be done with the theorem of the integration by parts, that says:

`intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx`.

We have to remember also the double-angle formula of sinus, that says:

`sin2alpha=2sinalphacosalpha`.

So, the integral becomes:

`2int_0^pie^costsintcostdt=-2int_0^pie^cost(-sint)*costdt=(1)`

We can assume that:

`f(t)=cost` and `g'(t)dx=e^cost(-sint)dt`.

So:

`f'(t)dt=-sintdt`

`g(t)=e^cost` and this is because `inte^f(x)f'(x)dx=e^f(x)+c`.

Then:

`(1)=-2{[cost(e^cost)]_0^pi-int_0^pie^cost(-sint)dt}=`

`=-2[coste^cost-e^cost]_0^pi=`

`=-2(cospie^cospi-e^cospi-(cos0e^cos0-e^cos0))=`

`=-2(-e^-1-e^-1-e+e)=4/e`.