Integration By Parts

MAT 229, Spring 2021

Week 4

Supporting materials

If you wish to get a different perspective on the notes below, try either of the following textbook sections.

Stewart’s Calculus
	Section 7.1: Integration by Parts

Boelkins/Austin/Schlicker’s Active Calculus
	Section 5.4: Integration by parts

Integration techniques

Anti-differentiation is the reverse of differentiation. Anything we can learn about anti-differentiation must come from differentiation. They’re reflections of each other in some weird Alice-in-Wonderland looking-glass.

Power rule and Power rule: (d/(d x))((x^n))=n (x^(n-1)) ⟶ ∫(x^k)d x=(1/(k+1))(x^(k+1))+C if k≠-1.

Chain rule and Substitution: (d/(d x))(F(u(x)))=F'(u(x))u'(x) ⟶ ∫f(u(x))u'(x)d x=∫f(u)d u

What about the Product rule? What’s its mirror reflection?

We want an integration technique that comes from the product rule for differentiation.
	((u(x)v(x)))'=u'(x)v(x)+u(x)v'(x)
If the two sides above are equal as functions, so are their integrals: so integrating both sides produces
	u(x)v(x)=∫u'(x)v(x)d x+∫u(x)v'(x)d x
Rewrite this as
	u(x)v(x)-∫u'(x)v(x)d x=∫u(x)v'(x)d x
And there’s your new rule, with the mysterious name of “Integration by Parts” (rather than “the product rule backwards”).

Integration by parts

Here’s how I usually come at it (using functions f and g, which are my favorites for the project rule). 
Let’s say you don’t like the integral ∫f(x)g'(x) d x. You can’t think of an anti-derivative of f(x)g'(x) . So rewrite it as 
					f(x)g(x)-∫f'(x)g(x) d x 
and maybe that one will look better to you (replace one pesky integral for another): you might recognize the integrand f'(x)g(x) , and know an anti-derivative.

Note that this works for definite integrals, too: we simply add limits:

(∫_a^b)f(x)g'(x)dx = f(x)g(x)|({{b}, {a}})-(∫_a^b)f'(x)g(x) d x

So you've got choices. If one looks better, do that one!
Here’s the key: look for  a product in the integrand, of two functions: one you wouldn't mind differentiating (f(x)), and the other you wouldn't mind anti-differentiating (think of it as g'(x)). And maybe you’ll get a product in the integrand that you like better.

Integration by parts -- alternate version

We should mention that some folks write integration by parts in a slightly different way: If you can identify u(x) and d v(x) in an integral, try rewriting the integral as
	∫u(x)dv(x)=u(x) v(x)-∫v(x)du(x)
This comes right out of the forms above, since we can think of the product 
	v'(x)dx = (dv/dx)dx =dv(x)
so
	∫u(x)v'(x)d x=∫u(x)dv(x)=u(x) v(x)-∫v(x)du(x)
It's really just a shorthand for what we've got above (and which I prefer).

People often suppress the x dependence: if you start with my favorite form, and if we write u=f(x) and dv=g'(x)dx, then du=f'(x)dx and v=g(x); and if we can identify an integral as

∫u d v = u v-∫v d u.

This makes it all look a little like a double substitution. It’s actually just a good shorthand. I personally prefer the first form we considered, above -- but you’re welcome to use this alternative form (and sometimes I do!).

Example

To evaluate ∫x (e^x)d x let’s try integration by parts. Identify u(x) and v'(x).
	u(x)=x
v'(x)=(e^x)
This leads to
	u'(x)=1
v(x)=∫v'(x)d x=∫(e^x)d x=((e )^x)
Only one antiderivative is needed; any value of C at this point in “+C” works. Applying the integration by parts formula ∫u d v=u v-∫v d u produces
	∫x (e^x)d x=x (e^x)-∫(e^x)d x
The new integral is one of our basic integrals. The final answer is
	∫x (e^x)d x=x (e^x)-(e^x)+C

Alternate form: Choose u=x and d v=((e )^x)d x; then du=dx and  v=((e )^x), so that the process looks like this:
	∫x (e^x)d x=∫u d v=u v-∫v d u=x (e^x)-∫(e^x)d x =x (e^x)-(e^x)+C
The same thing, of course! Just a different approach, which some folks like better.

Questions

Consider the integral ∫x cos(2x)d x.

I see a product, x and cos(2x). I don’t mind differentiating or anti-differentiating either one of these function. However, if I differentiate x it becomes 1; that will make for a nicer product. That suggests I choose f(x)=x, and g’(x)=cos(2x).

What is ∫x cos(2x)d x if you use integration by parts with u=x and d v=cos(2x)d x?

What would happen if you try to use integration by parts with u=cos(2x) and d v=x d x?

(Video)

Question

How can you use integration by parts to evaluate ∫(x^2)sin(3x)d x? (Video)

Question

How can you use integration by parts to evaluate ∫(tan^(-1))(x)d x? (Video)

Guidelines

Guidelines for choosing u and d v

You have to be able to integrate d v (that is, know the anti-derivative v).

The derivative of u and the integral of d v can’t get too much messier than they already are. You want ∫v d u to be no worse than ∫u d v.

Questions

To evaluate ∫(x^n) ln(x)d x, use integration by parts with u=ln(x) and d v=(x^n) d x. (Video)

intn[n_]:=Integrate[x^n Log[x],x]

Using your results from the previous question, what is ∫ln(x)d x? (Video)

intn[0]

Evaluate (∫_0^π)(x^2) (e^(-4x))d x using integration by parts. (Video)

Integrate[x^2 E^(-4x),{x,0,Pi}]
N[%]

Evaluate ∫(x^3) cos((x^2))d x using integration by parts, but first make a substitution. (Video)

Integrate[x^3 Cos[x^2],x]

What is the area of the region bounded by y=(sin^(-1))(x/2), the x-axis, and x=1? (Video)

Integrate[ArcSin[x],{x,0,1}]

What is the volume of the solid obtained by rotating about the x-axis  the region bounded by y=x ln(x) and the x-axis for 1≤x≤e. (Video)

Integrate[Pi (x Log[x])^2,{x,1,E}]

In the first question above, there is one special case: n=-1. Use integration by parts in this particular instance to get ∫(x^(-1))ln(x)d x=stuff-∫(x^(-1))ln(x)d x. Solve this equation for ∫(x^(-1))ln(x)d x to finish evaluating the integral (what a great trick!). (Video)

Integrate[ x^(-1)Log[x],x]

To evaluate ∫(e^x) cos(x)d x, use integration by parts twice. (Be sure to choose u and d v the same way both times. If you choose u=(e^x) the first time, be sure to choose u=(e^x) the second time. Or, if you choose u=trig the first time, choose u=trig the second time.) Then employ what you did in the last problem to finish evaluating the integral. (Video)

Integrate[ E^x Cos[x],x]

Homework

IMath problems on integration by parts.