Calculus Evaluating the indefinite integral?

Question

Evaluate the indefinite integral of:

e^(6x)sin(5x)dx

Can anyone show me how to do this one?

Answer 1

do two integration by parts with u = e^6x both times, then collect the two integrals on one side and divide by their coefficient

Answer 2

Integration by parts:∫u*dv=uv-∫v*du

J=∫e^(6x)sin5x dx=(1/6)∫sin5x d[e^(6x)]=

(1/6)e^(6x)sin5x-(1/6)∫e^(6x) d(sin5x)=

(1/6)e^(6x)sin5x-(5/6)∫e^(6x)cos5x dx=

(1/6)e^(6x)sin5x-(5/36)∫cos5x d[e^(6x)]=

(1/6)e^(6x)sin5x-(5/36)e^(6x)cos5x+

(5/36)∫e^(6x) d(cos5x)=

(1/6)e^(6x)sin5x-(5/36)e^(6x)cos5x-

(25/36)∫e^(6x)sin5x dx=

(1/6)e^(6x)sin5x-

(5/36)e^(6x)cos5x-(25/36)*J

J(1+25/36)=(1/6)e^(6x)sin5x-

(5/36)e^(6x)cos5x

J=(36/61)[(1/6)e^(6x)sin5x-

(5/36)e^(6x)cos5x]

J=(6/61)e^(6x)sin5x-(5/61)e^(6x)cos5x

J=[e^(6x)]*(6sin5x-5cos5x)/61