Try the substitution u = cos(x) instead and get the integral in terms of u and then note that you can integrate by parts.
Hey, could I get some guidance on how to do this question:
I figure you'd also have to use some trig identities here, correct?
My first approach was:
let u = e^(cos(x))
so du/dx = -sin(x)e^cos(x)
let dv/dx = sin(x)cos(x) = 0.5sin(2x)
so v = -0.25cos(2x)
=> uv - integral of -0.25cos(2x)e^cos(x)
.. and yay, we're back to square one practically
Integrate = I =cos(x)sin(x)e^(cos(x))
let u = cos(x)
du/dx = -sin(x)
du = sin(x)dx
=> -I = integral of (ue^u)du
let f = u let dg/du = e^u
df/du = 1 therefore g = e^(u)/u
ug - integral of (e^(u)/u) ????
which gives us e^u - e^(u)/u^2 (edit: I realise this is wrong)
= e^cos(x) - e^cos(x)/cos^2(x) + c
Still wrong, i believe, but i got further!
does this forums support latex?
where was my mistake?