# 11.2 Integration by parts

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

We can also integrate this expression,

As integration is linear^{1}^{1}
1
That is, the integral of a sum is the same as
the sum of the integrals., this is equivalent to

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

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

^{2}

^{2}2 Note that $\iff$ means \sayif and only if - i.e. that the two equations are equivalent.

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

###### Example 11.2.1

Find the value of

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

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!