The picture that made integration by parts stop feeling like a trick

For about a year I treated integration by parts like a spell. Pick $u$, pick $dv$, write the formula, hope. The answers came out right often enough. But if you had asked me why the formula is true, I would have said "product rule, something something", and that bothered me more than it should have. (Probably a personal problem.)

Then one Sunday I was avoiding a problem set and drew a graph just to look busy. It clicked. Here it is.

The formula, the boring way

$\int u\,dv = uv - \int v\,du$

You get it by differentiating a product, $d(uv) = u\,dv + v\,du$, integrating both sides and rearranging. That is a perfectly valid proof. It is also a proof that makes you feel nothing.

The picture

Forget $x$ and $y$ for a moment. Take a plane where the horizontal axis is $u$ and the vertical axis is $v$. These are just two quantities that both change as you walk along a curve. Say the curve starts at the origin and climbs to a point $(u, v)$, going up and to the right the whole way.

Draw the rectangle with corners $(0,0)$ and $(u,v)$. Its area is $uv$. Now the curve cuts that rectangle into two pieces:

• the piece under the curve, sitting on the $u$-axis. Slice it into thin vertical strips of height $v$ and width $du$. Its area is $\int v\,du$.
• the piece to the left of the curve, hugging the $v$-axis. Slice it into thin horizontal strips of length $u$ and height $dv$. Its area is $\int u\,dv$.

Those two pieces tile the rectangle with nothing left over. So

$\int u\,dv + \int v\,du = uv$

and moving one term across gives the formula. That is all it is. Integration by parts is "area of the whole rectangle minus the part you did not want". If the horizontal strips are painful to add up, add up the vertical strips instead and subtract.

This also explains the $[uv]$ term in definite integrals. It is the rectangle, evaluated at the two ends:

$\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du$

What it says about choosing $u$

The question is never "which letter is $u$". The question is: which way of slicing leaves me with an easier integral? The LIATE rule is just a ranking of which function becomes simpler when you differentiate it (that one goes in $u$) and which one stays friendly when you integrate it (that one goes in $dv$).

Example 1: $\int x e^x\,dx$

$u = x$, $dv = e^x dx$, so $du = dx$ and $v = e^x$.

$\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1) + C$

Check: $\frac{d}{dx}\big[e^x(x-1)\big] = e^x(x-1) + e^x = xe^x$. Good.

Why not the other way? With $u = e^x$ and $dv = x\,dx$ you get $\frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x\,dx$. The new integral is worse than the old one. Wrong slicing.

Example 2: $\int \ln x\,dx$, the one that looks impossible

There is no visible product. So make one: $\int \ln x \cdot 1\,dx$. Take $u = \ln x$, $dv = dx$, so $du = \frac{dx}{x}$, $v = x$.

$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C$

Definite version: $\int_1^e \ln x\,dx = \big[x\ln x - x\big]_1^e = (e - e) - (0 - 1) = 1$.

And here the rectangle gets a bonus. Take the curve $y = \ln x$ from $(1,0)$ to $(e,1)$. Under it is $\int_1^e \ln x\,dx = 1$. To its left (as a function of $y$, the curve is $x = e^y$) is $\int_0^1 e^y\,dy = e - 1$. Add them: $1 + (e-1) = e$, and the rectangle term is $uv$ at the end minus $uv$ at the start, $e\cdot 1 - 1\cdot 0 = e$. It matches. That is why the integral of an inverse function can be computed from the integral of the function.

Example 3: the one that eats itself

$I = \int e^x \sin x\,dx$. Do by-parts twice.

First: $u = \sin x$, $dv = e^x dx$:

$I = e^x\sin x - \int e^x\cos x\,dx$

Second, on the new integral: $u = \cos x$, $dv = e^x dx$, so $du = -\sin x\,dx$:

$\int e^x\cos x\,dx = e^x\cos x + \int e^x \sin x\,dx = e^x\cos x + I$

Put it back: $I = e^x\sin x - e^x\cos x - I$, so $2I = e^x(\sin x - \cos x)$ and

$I = \frac{e^x}{2}(\sin x - \cos x) + C$

One rule that saved me: in the second step, take the same type of function as $u$ that you took in the first step (trig both times). If you switch halfway, you undo your own work and end up with $I = I$, which is true but not useful.

Repeated by-parts: the table trick

For $\int x^2 e^x\,dx$ you would apply the formula twice. A faster way is to write two columns: keep differentiating $x^2$ until you hit zero, keep integrating $e^x$ alongside, and multiply diagonally with alternating signs.

sign	differentiate $x^2$	integrate $e^x$
+	$x^2$	$e^x$
−	$2x$	$e^x$
+	$2$	$e^x$
−	$0$	$e^x$

Multiply each entry in the middle column by the entry one row below it in the right column: $+x^2 e^x - 2x e^x + 2e^x$. So $\int x^2e^x\,dx = e^x(x^2 - 2x + 2) + C$. It only works when one function dies by repeated differentiation, but that is exactly the case that eats the most time in a mock.

Things that cost me marks

1. Losing the minus sign in front of $\int v\,du$ when $du$ itself is negative (like $\cos x$ above).
2. Forgetting to apply the limits to the $[uv]$ term before subtracting the second integral.
3. Not differentiating my answer at the end. It takes twenty seconds and catches nearly every sign error.

That is the post. If you only keep one thing: draw the rectangle once, by hand, and the formula stops being something you memorise.