i keep getting stuck and i think im missing something

Question

integrate by parts e^(bx) cos x dx

thanks

Answer 1

Integration by parts is basically using the formula
∫u(dv/dx)dx = uv - ∫v(du/dx)dx

In the case of I=∫f(x)dx where f(x)=e^(bx)cos(x)dx, you can apply repeated application of the formula (two times), to end up with a term -b^2∫f(x)dx on the right hand side. By moving this term to the left and factoring, you will get
I=e^bx(sin(x)+b*cos(x))/(1+b^2)

I will do the first step, and you can complete the integration:
Let
u=e^bx, (du/dx)=be^bx
v=sin(x), (dv/dx)=cos(x)
then
I=∫u(dv/dx)dx
=∫e^bx cos(x) dx

Using integration by parts,
I=uv - ∫v(du/dx)dx
=e^bx sin(x) - ∫sin(x)(be^bx)dx

Now apply Integration by parts again to the remaining integral on the right-hand-side, move the term b^2*I to the left, factor and solve for I, you should get the result shown above.

Differentiate the result (integral) to make sure you get back the original expression for f(x).