How do you integrate $\int e^{cos} (t) (sin 2 t) d t$ $inte^(cos)(t) (sin 2t) dt$ between $a = 0$ $a = 0$ , $b = π$ $b = pi$ ?

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Answer 1 · 2015-05-14T08:57:08

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:

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

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

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

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

How do you integrate inte^(cos)(t) (sin 2t) dt∫ecos(t)(sin2t)dtinte^(cos)(t) (sin 2t) dt between a = 0a=0a = 0, b = pib=πb = pi?