How do you integrate by parts?

When you are faced with an integral which is a product, such as x.cos(x), you may be able to integrate it by parts. The statement of the integration by parts is that:

∫u(dv/dx)dx = uv-∫v(du/dx)dx

So if you have a function of the form u(dv/dx) (such as x.cos(x)​) it can be presented in the above form. In this case u = x and dv/dx = cos(x). 

When doing an intergration by parts it is useful to draw a grid first and work out what v and du/dx are (since we already know u and dv/dx) like so:

u = x

dv/dx = cos(x)

v = sin(x) (this is what you get when you integrate cos(x) as ∫(dv/dx)dx = v)

du/dx = 1 (what you get from differentiating u)

And so now that we have everything we need we can plug things into the equation:

∫x.cos(x)dx = uv-∫v(du/dx)dx = x.sin(x) - ∫1.sin(x)dx

and then to finish we integrate the last bit:

 ∫x.cos(x)dx = x.sin(x)+cos(x)+c (since this is an indefinite integral we must add a constant of integration c).

TW
Answered by Tim W. Maths tutor

6342 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)


Use logarithms to solve the equation 2^(n-3) = 18000, giving your answer correct to 3 significant figures.


Consider the functions f and g where f (x) = 3x − 5 and g (x) = x − 2 . (a) Find the inverse function, f^−1 . (b) Given that g^−1(x) = x + 2 , find (g^−1 o f )(x) . (c) Given also that (f^−1 o g)(x) = (x + 3)/3 , solve (f^−1 o g)(x) = (g^−1 o f)(x)


Evaluate the following integral: (x^4 - x^2 +2)/(x^2(x-1)) dx


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning