# Integration by Parts and the Constant of Integration [duplicate]

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The constant of integration only seems to be used at the very end of integration by parts despite the use of integrals beforehand.

An example of this would be: $$\int x\sin(x)\ dx = x\int sin(x)\ dx - \int x'(\int sin(x)\ dx)\ dx$$ Ordinarily, the right side of the equation would be simplified to: $$x(-cos(x)) - \int-cos(x)\ dx$$ And further to: $$-x(cos(x)) + sin(x)$$ Then finally arranged and given the constant of integration: $$sin(x) - xcos(x)+ C$$

What I am confused about is why $C$ is only added at the very end of this instead of at each integral.

I would be more inclined to use try something more like this: $$x(-cos(x) +C_1) - \int -cos (x) + C_2\ dx$$ Which would simplify to: $$-x(cos(x) -C_1) - \int-cos(x)\ dx + \int C_2\ dx$$ And further to: $$-xcos(x) +C_1x +sin(x)+ C_3 + C_2x + C_4$$ Which finally arranges itself as: $$sin(x) - xcos(x) + C_5x + C_6$$ Where $C_5=C_1 + C_2$ and $C_6=C_3 + C_4$

I also feel I should probably mention I am a bit of an oblivious idiot so if the answer is completely obvious or my math is full of errors, I apologize.

## marked as duplicate by Zain Patel, Hans Engler, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 17 '16 at 4:10

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• If you use $-\cos x+C_1$ in the first term, then you must use $\int (-\cos x+C_1)\,dx$ in the second integral. After you integrate, you will notice that the $C_1$'s cancel. The integration by parts formula, in brief, says $\int u\,dv=uv-\int v\,du$. If you use $-\cos x+C_1$ for $v$, when applying the formula you need to use that $v$ wherever $v$ occurs. – André Nicolas Jul 16 '16 at 23:58
• @AndréNicolas I see now, I got it mixed up in my head that for each use of $v$ there was its own individual $C$. – N.D.H. Jul 17 '16 at 0:18
• Another way of looking at it: it is a theorem that if you know a specific antiderivative of some function, then every antiderivative of that function differs from your specific one by an additive constant $C$. (This followed from the fact that only constant functions have zero derivative everywhere, which is itself a consequence of the mean value theorem, say.) So you can ignore the "constant of integration" throughout the entire procedure (setting it to $0$ each time, for instance), end up with a specific antiderivative, and then apply the theorem to add $+C$ to the end. – Greg Martin Jul 17 '16 at 1:20
• the integration by parts really says that $(uv)' = u'v+ v'u$, integrating both side on $[a,b]$ : $\int_a^b (u(x)v(x))' dx = u(b)v(b)-u(a)v(a)= \int_a^b u'(x) v(x) dx + \int_a^b u(x) v'(x) dx$ or $$\int_a^b u(x) v'(x) dx = u(b)v(b)-u(a)v(a) -\int_a^b u'(x) v(x) dx$$ with more constant added, you can rewrite it $((u+C)(v+D))' = u'(v+D)+ v'(u+C)$ that leads to $$\int_a^b (u(x)+C) v'(x) dx = (u(b)+C)(v(b)+D)-(u(a)+C)(v(a)+D) -\int_a^b u'(x) (v(x)+D) dx$$ – reuns Jul 17 '16 at 2:29

## 1 Answer

Great questions. Indeed you are right that constents need to be added along the way but as you'll see, the algebra works out that only one remains in the end.

$\int f(x)g'(x) dx$ = f(x) (g(x)+c) - $\int f'(x) (g(x)+c ) dx$

thus

$\int f(x)g'(x) dx$ = f(x)g(x) + cf(x) - ($\int f'(x)g(x)dx$ + $\int cf'(x) dx$)

since our c is a constant we have that

$\int f(x)g'(x) dx$ = f(x)g(x) + cf(x) - $\int f'(x)g(x)dx$ - cf(x)

and finally we have our desired result that:

$\int f(x)g'(x) dx$ = f(x)g(x) - $\int f'(x)g(x)dx$

note that your C1 and C2 are in fact the same as they are the constants that appear when taking the anti derivative of g'(x).

($int\ g'(x) dx$ = g(x) + C)

with respect to your C3 and C4, we can simply refer to the sum of the two as "c" which we add in the end of the integration.