How do you integrate by parts?

To integrate the product of two functions - say f(x) = xsinx, the product of g(x) = x and h(x) = sin(x) - the we use integration by parts. 

Since integration is the inverse of differentiation, we'll appeal to the Product Rule for differentiation to find a rule for integration. The Product Rule says that (fg)' = f'g + g'f. Rearranging giives:

f'g = (fg)' - g'f.

Integrating we have:

Integral(f'g) = Integral((fg)') - Integral(g'f), or:

Integral(f'g) = fg - Integral(g'f).

This is integration by parts. We designate one of the functions to be f' and one to be g. You can see on the far right we have another integral! For the method to be of use, we need to make the new integral simpler (or at least no more complicated) than the one we began with. So, we want g'f to be simpler then f'g: in other words, when we assign functions f' and g we want g to be the one that gets simpler when differentiated. Let's try an example: let's try and integrate xsinx dx. First, we will choose g to equal x, since x becomes 1 under differentiation, but sinx goes to cosx.

I = Integral(xsinx) = xIntegral(sinx) - Integral(x'Integral(sinx))

I = -xcosx - Integral(-cosx)

I = sinx - xcosx +C

So as long as we choose f' and g with some forethought, it's easy enough to integrate products of functions. Note: Ususally we denote the rule with f' = dv and g = du, so 

Integral u dv  = uv - Integral v du.

This is just a fancy way of stating what we've already established.

Answered by Patrick K. Maths tutor

5205 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

make into a cartesian equation= x=ln(t+3) y= 1/t+5


Find the exact solution, in its simplest form, to the equation 2ln(2x+1) - 10 = 0.


How can the y=sin(x) graph be manipulated?


The line AB has equation 5x + 3y + 3 = 0. The line AB is parallel to the line with the equation y = mx + c. Find the value of m.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences