11.2 Integration by parts ‣ Chapter 11 Integral calculus ‣ Part I A Level mathematics (and further mathematics) ‣ A calculus of the absurd‣ Chapter 11 Integral calculus ‣ Part I A Level mathematics (and further mathematics) ‣ A calculus of the absurd

When differentiating a function $f(x)=a(x)\cdot b(x)$ , the product rule (as in Section 10.5) tells us that

\frac{d}{dx}\left[f(x)\right]=\frac{d}{dx}\left[a(x)\right]b(x)+a(x)\frac{d}{% dx}\left[b(x)\right]

(11.1)

We can also integrate this expression,

\int\frac{d}{dx}\left[f(x)\right]dx=\int\left(\frac{d}{dx}\left[a(x)\right]b(x% )+a(x)\frac{d}{dx}\left[b(x)\right]\right)dx

(11.2)

As integration is linear¹¹ 1 That is, the integral of a sum is the same as the sum of the integrals., this is equivalent to

\int\frac{d}{dx}\left[f(x)\right]dx=\int\frac{d}{dx}\left[a(x)\right]b(x)dx+% \int a(x)\frac{d}{dx}\left[b(x)\right]dx

(11.3)

As (by definition) $\int\frac{df}{dx}dx=f(x)$ , it is the case that

f(x)=a(x)\cdot b(x)=\int\frac{d}{dx}\left[a(x)\right]b(x)dx+\int a(x)\frac{d}{% dx}\left[b(x)\right]dx

(11.4)

Why is this useful? First we can do a little addition/rearranging

	$\displaystyle a(x)\cdot b(x)=\int\frac{d}{dx}\left[a(x)\right]b(x)dx+\int a(x)% \frac{d}{dx}\left[b(x)\right]dx$		(11.5)
	$\displaystyle\iff\int\frac{d}{dx}\left[a(x)\right]b(x)dx=a(x)\cdot b(x)-\int a% (x)\frac{d}{dx}\left[b(x)\right]dx$		(11.6)

This gives us (a somewhat awkward) way to integrate products of expressions!

First, fear not the fact that this is an integral in $z$ . As with sums, the variable itself does not matter (we could rename $z$ to anything else, and the integral would still be the same integral)! Note that we could solve this without using integration by parts; we could think about what would differentiate to give us $z\sin(z)$ , but this is no fun (it’s much nicer to have a systematic way to solve a problem than to rely on lucky guessing)!

We start by trying to work out what should be what in the identity in Equation 11.6. We are starting with the left-hand side, and want to set $\frac{d}{dz}\left[a(z)\right]=\sin(z)$ and $b(z)=z$ . Why? Because in the integral on the right-hand side, we calculate the integral of the integral of $a(z)$ and the derivative of $b(z)$ and what makes the problem \sayhard is not that we don’t know how to find the integral of either $\sin(z)$ or $z$ , it’s that we don’t know how to find the integral of the two multiplied together! As the derivative of $z$ is just $1$ , this gives us a way to rewrite the integral of the product of $z$ and $\sin(z)$ in terms of only the integral of $\sin(z)$ .

Let us proceed by using the formula for integration by parts (let us agree to set $x=z$ if you’re not convinced that the variables can be renamed without changing the structure of the expressions), that is

\int z\sin(z)dz=\left(\int\sin(z)dz\right)z-\int\left(\int\sin(z)dz\right)dz

(11.8)

The $\int\sin(z)dz$ appears, because if $\frac{d}{dz}\left[a(z)\right]=\sin(z)$ , then $a(z)=\int\sin(z)dz$ . We can proceed by simply evaluating the integrals on the right-hand side (all of which we know how to integrate)!

$\displaystyle\int z\sin(z)dz$	$\displaystyle=-\cos(z)z-\int\left(-\cos(z)\right)dz+C$	(11.9)
	$\displaystyle=-\cos(z)z+\int\cos(z)dz+C$	(11.10)
	$\displaystyle=-\cos(z)z+\sin(z)+C$	(11.11)

That this is the correct integral can be verified by differentiating the expression. You might also ask, why does the constant of integration $C$ remain unchanged after each step (doesn’t each integral add something to it). The answer is that it doesn’t - at each step we are effectively redefining $C$ to be $C_{\text{new}}=C_{\text{old}}+k$ (where $k$ is the constant of the new integral) - if we were to create a new constant each time it would get messy, so we can just \sayfold all the constants into one new super-constant!

11.2 Integration by parts

Example 11.2.1