MAT187 L16 - Integration By Parts

#MAT187 PCE
Recall from MAT186 that product rule for two differentiable functions is:

(u (x) v (x))^{'} = u^{'} (x) v (x) + u (x) v^{'} (x)

We can take this trick to derive the integration by parts formula. First, integrate both sides with respect to x:

\int (u (x) v (x))^{'} d x = \int u^{'} (x) v (x) d x + \int u (x) v^{'} (x) d x

The first integral cancels out the derivative inside it (they are inverse functions of each other), so:

u (x) v (x) = \int u^{'} (x) v (x) d x + \int u (x) v^{'} (x) d x

And rearranging that, we get:

\int u (x) v^{'} (x) d x = u (x) v (x) - \int u^{'} (x) v (x) d x

If we express $u (x)$ as $u$ , $v (x)$ as $v$ , $d v = v^{'} (x) d x$ , and $d u = u^{'} (x) d x$ we get:

\int u d v = u v - \int v d u

Integration by Parts

Given differentiable functions $u$ and $v$ :

\int u d v = u v - \int v d u

Integration by parts helps when one of the functions is infinitely differentiable, but the other is finitely differentiable.

Example: Integration by Parts of $\int x e^{2 x} d x$ .

Let $u = x$ , and $d v = e^{2 x} d x$ . Then,

\begin{aligned} \int x e^{2 x} d x & = x (\frac{1}{2} e^{2 x}) - \int \frac{1}{2} e^{2 x} d x \\ = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C \end{aligned}

Alternatively, if we let $u = e^{2 x}$ , and $d v = x d x$ . Then,

\int x e^{2 x} d x = \frac{1}{2} x^{2} e^{2 x} - \int \frac{1}{2} x^{2} e^{2 x} d x

...which does not help us because the integral becomes more complicated.
We chose $u = x$ because $x$ is finitely differentiable, whereas $e^{2 x}$ is infinitely differentiable.

Example: Integration by Parts of $\int x^{3} e^{x^{2}} d x$

(From question 39 of LEC101 Slides)
For this question, there are many options for $u$ and $d v$ . Not all of them will lead to useful values of $d u$ and $v$ , however.

u	dv	v	du
$x^{3}$	$e^{2 x^{2}}$	—	$3 x^{2}$
$x^{2}$	$x e^{x^{2}}$	—	$2 x$
$x$	$x^{2} e^{x^{2}}$	—	1
$1$	$x^{3} e^{x^{2}}$	—	0
$e^{x^{2}}$	$x^{3}$	$\frac{1}{4} x^{4}$	$2 x e^{x^{2}}$
$x e^{x^{2}}$	$x^{2}$	$\frac{1}{3} x^{3}$	$e^{x^{2}} + x (2 x) e^{x^{2}}$
$x^{2} e^{x^{2}}$	$x$	$\frac{1}{2} x^{2}$	$2 x e^{x^{2}} + x^{2} (2 x) e^{x^{2}}$
$x^{3} e^{x^{2}}$	$1$	$x$	$3 x^{2} e^{x^{2}} + x^{3} (2 x) e^{x^{2}}$
Side note: the in class method for solving this is a little different from what I did. Oh well.

We can use integration by substitution first to make the integral easier to evaluate.
Let $u = x^{2}$ . Then,

\frac{d u}{d x} = 2 x ⟹ d x = \frac{d u}{2 x}

Thus,

\int x^{3} e^{x^{2}} d x = \frac{1}{2} \int u e^{u} d u

...this allows us to use integration by parts much easier now.

\begin{aligned} \frac{1}{2} \int u e^{u} d u & = \frac{1}{2} (u e^{u} - e^{u}) + C \\ = \frac{1}{2} (x^{2} e^{x^{2}} - e^{x^{2}}) + C \\ = \frac{1}{2} e^{x^{2}} (x^{2} - 1) + C \end{aligned}

Example: Integration by Parts of $\int e^{x} \sin x d x$

(From question 40 of LEC101 Slides)
For this we can use integration by parts twice, since $\sin x$ differentiates back to itself after two iterations (with an added negative sign).
Let $u = \sin x$ , $d v = e^{x}$ . Then,

\begin{aligned} \int e^{x} \sin x d x & = e^{x} \sin x - \int \cos x e^{x} d x \\ = e^{x} \sin x - (\cos x e^{x} - \int - \sin x e^{x} d x) \\ = e^{x} \sin x - \cos x e^{x} - \int \sin x e^{x} d x \end{aligned}

Now, we can rearrange and isolate for $\int e^{x} \sin x d x$

\begin{aligned} 2 \int e^{x} \sin x d x & = e^{x} \sin x - \cos x e^{x} \\ = \frac{1}{2} (e^{x} \sin x - \cos x e^{x}) \\ = \frac{1}{2} e^{x} (\sin x - \cos x) \end{aligned}

Integration by Parts

Example: Integration by Parts of ∫xe2xdx.

Example: Integration by Parts of ∫x3ex2dx

Example: Integration by Parts of ∫exsin⁡xdx

Example: Integration by Parts of $\int x e^{2 x} d x$ .

Example: Integration by Parts of $\int x^{3} e^{x^{2}} d x$

Example: Integration by Parts of $\int e^{x} \sin x d x$