Integration by Parts

SCH_CALC L3CALC

Author: Sedecember
a3060100s@gmail.com

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When integrating, we might come across garbage integrals like

$$\int x{e}^{x}dx$$

Try u-sub; let $u=x$, $\frac{du}{dx}=1⟺dx=du$

$$⟹\int u\cdot e^{x}du$$

Doesn’t work. Try let $u=e^x$, $\frac{du}{dx}=e^x⟺dx=\frac{du}{e^x}$

$$⟹\int x\cdot u\frac{du}{e^x}$$ $$⟹\int xdu$$

still doesn’t work.

So, are we cooked?

Introduction

No, we are NOT cooked. We never are. To evaluate shitty integrals like this, mathematicians found a way called ‘Integration by Parts’!

This method comes from the product rule. Note that Juyoung has already proved the product rule from first principles here.
The product rule says:

$$[uv{]}^{\prime }=u^{\prime }v+v^{\prime }u$$

Where $u$ and $v$ are functions in terms of $x$.

But… we can manipulate the product rule further by adding integrals on both sides like this

$$\int [uv{]}^{\prime }dx=\int u^{\prime }vdx+\int v^{\prime }udx$$ $$⟹uv=\int u^{\prime }vdx+\int u{v}^{\prime }dx$$

Now, if we manipulate this further, we get

$$⟹\boxed{\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx}$$

That boxed expression is the formula for ‘integration by parts’.

Application (Indefinite Integrals)

The question is now, how do we use this?

This is used just like the product rule in differentiation. You just select which function serves as $u$, and which serves as $v$.

The method I use is labelling each individual $u$, $u^{\prime }$, $v$, $v^{\prime }$.

Lets start with the shitty integral from above

$$\int x{e}^{x}dx$$

For LHS, we need to have $u$ and $v^{\prime }$. If we have those, then we can calculate $u^{\prime }$ and $v$ for RHS.

Let’s start with $u$ and $v^{\prime }$:
$u=x$
$v^{\prime }=e^x$
…Now this implies:
$u^{\prime }=1$
$v=e^x$

Now let’s set up the integral

$$\int x{e}^{x}dx=x{e}^x-\int 1\cdot e^{x}dx$$ $$\int x{e}^{x}dx=x{e}^x-e^x+C=\boxed{e^x(x-1)+C}$$

Application (Definite Integrals)

Does it serve differently for definite integrals?

Yes, but not much.

All it changes is that we also apply the bounds to $uv$.

$$⟹\boxed{{\int }_a^{b}u{v}^{\prime }dx=[uv{]}_a^b-{\int }_a^b{u}^{\prime }vdx}$$

That’s it. Just some simple integrations from then!