Is Integration by Parts Incorrect for ∫(x2 + 7x) cosx dx?

[loserspearl]

∫(x² + 7x) cosx dx

If I make v = (x² + 7x) and du = cosx dx I get
((x² + 7x) sinx)/2

If I make v = cosx and du = (x² + 7x) dx I get
((x³/3 + 7x²/2) cosx)/2

using the form X=Y-X to X=Y/2

Neither are correct, what did I do wrong?

 

[jz92wjaz]

Integration by parts:
∫u dv = uv - ∫v du

You are doing the integration by parts incorrectly.

 

[SteamKing]

∫(x² + 7x) cosx dx

If I make v = (x² + 7x) and du = cosx dx I get
((x² + 7x) sinx)/2

If I make v = cosx and du = (x² + 7x) dx I get
((x³/3 + 7x²/2) cosx)/2

using the form X=Y-X to X=Y/2

Neither are correct, what did I do wrong?

I recommend you expand the integrand (x² + 7x) cosx into x² cos x + 7x cos x, and integrate the two resulting terms.
I would start with integrating the second term of this expansion first.

 

[loserspearl]

K thanks