how do you integrate e^(sinx)sin2x

Question

it is valued from 0 to pi/2

Answer 1

Sin(2x)=2sin(x)cos(x).

Let u=sin(x); du=cos(x)dx.

S[0,(pi)/2](e^sin(x)sin2xdx) denotes the integral in the species range.

This can now be written: S[0,1](e^u.2udu)=2S[0,1](ue^udu).

Let p=u, dp=du; dq=e^udu, q=e^u. d(pq)=qdp + pdq; d(ue^u)=e^udu+ue^udu.

Integrating: ue^u=e^u+S(ue^udu); 2S[0,1](ue^udu)=2e^u(u-1)[0,1]=2(0-(-1))=2.

Answer 2

Expand sin(2x)

= 2(integral)sin(x)cos(x)[e^cos(x)]

Let u=sin(x); du=cos(x)dx.

int e^u2u du from 0 to 1

= 2int e^u*u du from 0 to 1

Let p=u, dp=du

dq=e^udu, q=e^u

d(pq)=qdp + pdq

d(ue^u)=e^udu+ue^udu

int ue^u = e^u+S(ue^udu)

2ue^u du from 0 to 1

2e^u(u-1) from 0 to 1

2(0-(-1))=2

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