When we integrate by parts, for example as in $\displaystyle \int{t\sin{t} }dt$ if $u = t$ and $dv = \sin{t}dt$ then we get $$\displaystyle \int{t\sin{t} }dt = t \displaystyle \int{\sin{t}}dt-\int{\sin{t}dt} = t(-\cos{t}+C_1)+\cos{t}+C_2 = -t\cos{t}+tC_1+\cos{t}+C_2$$ as opposed to $$\displaystyle \int{t\sin{t} }dt = -t\cos{t}+\cos{t}+C.$$

Are these two forms the same? If not, where is my error?

  • 1
    $\begingroup$ You should not have $C_1$. When you find $v$, you are not computing the indefinite integral of $dv$, meaning the set of all functions whose differential is $dv$, but instead you are computing one function whose differential is $dv$. $\endgroup$ – bartgol Mar 14 '16 at 21:56
  • $\begingroup$ An argument 'a posteriori' is that any two antiderivatives should differ by a constant, which implies $C_1=0$. $\endgroup$ – bartgol Mar 14 '16 at 21:56
  • $\begingroup$ @bartgol I don't understand your first argument. $\endgroup$ – Puzzled417 Mar 14 '16 at 21:57

Integration by parts says that:

$$\int f'(t)g(t)\,dt = f(t)g(t)-\int f(t)g'(t)\,dt$$

$g(t)=t$ and $f'(t)=\sin t$. But if you choose $f(t)=\cos t+C_1$, you have to use that same $f(t)$ in every place:

$$\int t\sin t\,dt = (\cos t+C_1)t - \int (\cos t+C_1)\,dt$$

(You've got an error in your integral, with $\sin t$ on the right side.)

As you can see, the two $C_1$s cancel out.

  • $\begingroup$ So we to keep things simple just say $C_1 = 0$? $\endgroup$ – Puzzled417 Mar 14 '16 at 22:04
  • $\begingroup$ Yes, there is never a reason, in integration by parts, to be concerned about the multiple possible anti-derivatives. As phrased in terms of $f,g$, you see that having any $f,g$ that work is fine, so we take the easiest. $\endgroup$ – Thomas Andrews Mar 14 '16 at 22:06

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