LinkBack URL
About LinkBacksBecause it is not necessary. We use $+c$ only to say that any constant may work.Originally Posted by Foreverlearning
Integration By Parts Help
Hey Guys,
Can someone please explain to me why the constant isn't used in the v of the u●v part of the formula?
I.e. Integral u●dv = integral u●v - integral v●du
For example Integral x cos (x) dx
Let x = u and dv = cos(x)dx, then du = 1 and v = sin(x)+c1
=> Integral x cos (x)dx = x sin(x)+c1 - Integral sin(x)+c1 dx
= x sin(x)+c1 - (Integral sin(x)dx + Integral c1 dx)
= x sin(x)+c1 - (cos(x) + c2 + c1x +c2))
= x sin(x)+c1 + cos(x) - c2 - c1x - c2
=> x sin(x) +c1 + cos(x) - c2 - c1x - c2 = x sin(x) +c1 + cos(x) - c1x - 2c2 .
Now the generic solution is xsin(x) + cos(x) + c.
What am I doing wrong?
Thanks.
Thanks for the fast reply,
I realise that it is not necessary but could you please show me mathematically why this requirement is not necessary?
Could you please show how the long-winded mathematical calculation is equivalent to the shorter methods values?
I am having trouble wrapping my head around how two different constant values and one of those values also multiplying the variable equates to the same value.
I look forward to your reply.
Consider the two functions $y(x)=x\sin(x)+\pi~\&~z(x)=x\sin(x)-\pi$. Clearly they are different functions.
But they have the same derivative, $y'(x)=z'(x)$
So the anti-derivative of $\sin(x)+x\cos(x)$ is unique up to a constant: $x\sin(x)+c$
  |
