Section 7.1: Integration By Parts Worksheet

Integration by parts

Let’s say you don’t like the integral ∫f(x)g'(x) d x. You can rewrite it as 
f(x)g(x)-∫f'(x)g(x) d x 
if you like, and maybe that one will look better to you (one integral for another).

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. The first thing to look for in the integrand is a product, 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)).

Alternate form

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, then we can rewrite it as 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!).

Questions to submit

To evaluate ∫(x^n) ln(x)d x, use integration by parts with f(x)=ln(x) and g'(x)=(x^n).

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

Using your results from problem one, what is ∫ln(x)d x?

intn[0]

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

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.

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

What is the area of the region bounded by y=(sin^(-1))(x), the x-axis, and x=1/2? (How did we find the anti-derivative of (tan^(-1))?)

Integrate[ArcSin[x],{x,0,1/2}]

What is the volume of the solid obtained by rotating about the x-axis  the region bounded by y=x (Sqrt[ln(x)]) and the x-axis for 1≤x≤e. (Use problem one!)

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

In problem one 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. (Can you think of how this becomes a general rule? What is special about the integrand,  (x^(-1))ln(x)?)

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 problem 7 to finish evaluating the integral.

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