Why Ignore the Integration Constant in ∫xcos(x) dx?

[MHD93]

Hi

When we use the integration by parts to find the integral of xcosx dx, we assume that cosx dx = du and integrate both sides to find u
When we integrate
du = cosxdx
we find that u = sinx
the question is why don't we take the integration constant C into account?

 

[Petr Mugver]

the question is why don't we take the integration constant C into account?

You can take it into account if you want, you'll get the same result (except for a final integration constant of course), so it's usually better to assume this constant zero in the intermediate steps.

 

[MHD93]

so it's usually better to assume this constant zero in the intermediate steps.

But it may or mayn't be zero, therefore the final integration is different if it's not zero.

 

[MHD93]

I indeed am in need of your help

 

[phyzguy]

The point is that it cancels out. Without the integration constant:
\int{udv} = uv-\int{vdu}
With the integration constant:
\intudv = u(v+C)-\int{(v+C)du} = uv+uC-\int{vdu}-C\int{du} = uv +uC-\int{vdu}-uC = uv-\int{vdu}

 

[MHD93]

Wow, that's real helpful, thank you

 

[paulfr]

Hi

When we use the integration by parts to find the integral of xcosx dx, we assume that cosx dx = du and integrate both sides to find u When we integrate du = cosxdx we find that u = sinx

For this integrand it is much better to let u = x and dv = Cos x
Your choice will work too, but if x is raised to a power the clear choice is
u = x^n and dv = Cos x
This integral is also best done with Tabular Integration By Parts