# Integrating by parts + substitution

• Nov 3rd 2012, 11:21 PM
99.95
Integrating by parts + substitution
Hey, could I get some guidance on how to do this question:

Integrate cos(x)sin(x)e^(cos(x))

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 :(
• Nov 4th 2012, 12:39 AM
chiro
Re: Integrating by parts + substitution
Hey 99.95.

Try the substitution u = cos(x) instead and get the integral in terms of u and then note that you can integrate by parts.
• Nov 4th 2012, 01:11 AM
99.95
Re: Integrating by parts + substitution
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

again:
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?
• Nov 4th 2012, 01:52 AM
Prove It
Re: Integrating by parts + substitution
Quote:

Originally Posted by 99.95
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

again:
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?

Last time I checked, the integral of e^u is e^u, not (e^u)/u...
• Nov 4th 2012, 01:57 AM
TheSaviour
Re: Integrating by parts + substitution
If dg/du = e^u then g = e^u, not [e^(u)]/u.

EDIT: yeah, like Prove It said.