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For context, the following is the cyclic integration by parts problem: $$ I = \int \sin (2x) \cos (3x) dx $$

After first setting $u$ as $\sin (2x)$ and $dv$ as $\cos (3x)$, I get the following expression:

$$ I = \frac{1}{3} \sin(3x)\sin(2x) - \frac{2}{3}\int \sin(3x)\cos(2x) dx $$ At this point, when I set $u$ as $\sin(3x)$ and $dv$ as $\cos(2x)$, I get the intermediate form of $I=I$, while reversing choices for $u$ and $dv$ yields the correct expression of: $$ \frac{3}{5}\sin(2x)\sin(3x)+\frac{2}{5}\cos(3x)\cos(2x) $$ Is there any specific reason for why the choice matters. The only implication that I can think of is the the two functions are not commutative; however, this does not seem to be the reason as these two trig functions are generally commutative.

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    $\begingroup$ The 1st time, you differentiated $\sin2x$ to get $\cos2x$ (actually, $2\cos2x$, but for our purposes here, the constant multiplier is irrelevant). Now if the 2nd time you set $dv=\cos2x\,dx$, then you're undifferentiating $\cos2x$; you're undoing what you did the first time, so of course you get back to where you started from, and $I=I$. $\endgroup$ Feb 16, 2022 at 1:50

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It is important to remember that the u-v substitutions for "Integration by Parts" comes from the following identity for derivatives: $$ \Big(U(x)\cdot V(x)\Big)' \ = \ U(x)\cdot V'(x) \ + \ U'(x) \cdot V(x) $$ If you integrate both sides of that expression, and remember that $\int{F'(x) dx}=F(x)$ simplifies the left hand side, you are left with: $$ U(x)\cdot V(x) \ =\int{U(x)\cdot V'(x) \ dx} \ + \ \int{U'(x)\cdot V(x) \ dx} $$ that rearranges into: $$ \int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \int{U'(x)\cdot V(x) \ dx} $$ which is the form we usually see the identity in.


When you repeat that process you have to be careful with your new change of variables, so lets call them $Y(x)$ and $Z(x)$ to be clear.

$$ \int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \int{Y(x)\cdot Z'(x) \ dx} $$

The "Integration by Parts" identity then gives us:

$$ \int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \Big( \ Y(x)\cdot Z(x)\ - \ \int{Y'(x)\cdot Z(x) \ dx}\Big) $$

or rather

$$ \int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ Y(x)\cdot Z(x)\Big) \ + \ \int{Y'(x)\cdot Z(x) \ dx} $$

which is all good and correct thusfar, but it doesn't tell us whether we chose $Y(x)=V(x)$ or $Y(x)=U'(x)$ for the substitution. If you choose $Y(x)=V(x)$ then $Z'(x)=U'(x)$ and you can maybe start to see why you're getting $I=I$:

$$ \int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ V(x)\cdot U(x)\Big) \ + \ \int{V'(x)\cdot U(x) \ dx} $$

It's like Gerry Myerson said in the comments, that choice of substitution is essentially the reverse of the previous step.

On the other hand, if $Y(x)=U'(x)$ then $Z'(x)=V(x)$ and so you get:

$$ \int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ U'(x)\cdot \int V(x)\ dx \Big) \ + \ \int{\Big(U''(x)\cdot \int V(x)\ dx \Big)\ dx } $$

which is actually making forward progress by taking deeper-and-deeper derivatives of $U(x)$ and higher-and-higher anti-derivatives of $V'(x)$.

(Swapping which variable gets the derivative midway through was essentially taking a derivative-then-antiderivative of $U(x)$ and an antiderivative-then-derivative of $V'(x)$... which is exactly why it all cancelled out.)

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Frequently it does not matter which factor you integrate and which you differentiate. It is important that once you make a decision stick with it.

That is:

$\int \sin (2x)\cos (3x)\ dx\\ u=\sin 2x, dv = cos 3x\ dx$
If we wanted to switch wihch term was $u$ and which was $dv$ it would not make a difference. Sticking with this start...
$du = 2\cos 2x\ dx, v = \frac {sin 3x}3\\ \frac{\sin 2x\sin 3x}{3} - \frac 23 \int \sin 3x\cos 2x\ dx$
Now it will make a difference. If we chose $u = \sin 3x, dv = \cos 2x\ dx $ we will be back where we started.

For an example of an integral where it makes a little bit more difference which factor is the $u$ and which is $dv$

$\int \sec^3 x\ dx\\ u = \sec x, v = \sec^2 x\ dx\\ du = \sec x\tan x, dv = \tan x\ dx\\ \sec x\tan x - \int sec x\tan^2 x\ dx\\ \sec x\tan x - \int (sec x)(sec^2 x - 1)\ dx\\ I = \sec x\tan x +\int \sec x\ dx - I\\ I = \frac 12 (\sec x\tan x + \ln|sec x + tan x|) + C$

By the way, in the original example, you can bypass integration by parts.

$\sin(2x)\cos(3x) = \sin(2x+3x) + \sin(2x-3x)$

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