How to integrate by parts

Sometimes while integrating, we may come across an expression that is not a polynomial, and thus we cannot use the convenient power rule to integrate. Consider the function y=xcos(x). It is not immediately clear how we should start we this one, however upon further inspection, we may introduce the technique of Integrating by Parts. Essentially we split the function into two parts, say u and v, and then employ a formula which allows us to integrate them together:

∫u·dv = u·v − ∫v·du

Applying this to our function, we obtain

∫x·cos(x) dx = u·v − ∫v·du

Here we note that integrating cos(x) is a lot simpler than integrating x, and differentiating x is also simpler than differentiating cos(x), so it would make sense to set

u = x and dv = cos(x)

This in turn gives us

du = 1 and v = sin(x)

Thus plugging these values back into our original formula, we get

∫x·cos(x)= x·sin(x) - ∫sin(x)·1

So now, all we need to do is integrate sin(x), which is definitely easier than what we started with. Thus, the end product gives us 

∫x·cos(x)= x·sin(x) + cos(x) + C

where C of course is the constant of integration.

Answered by Dennis L. Maths tutor

3193 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Express 2Cos(a) - Sin(a) in the form RCos(a+b) Give the exact value of R and the value of b in degrees to 2 d.p.


Find the coordinates of the sationary points on the curve x^2 -xy+y^2=12


Integrate (x+2)/((x+5)(x-7)) using partial fractions between the limits 5 and -2, giving your answer to 3sf


Three forces, (15i + j) N, (5qi – pj) N and (–3pi – qj) N, where p and q are constants, act on a particle. Given that the particle is in equilibrium, find the value of p and the value of q. (Mechanics 1 June 2017)


We're here to help

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

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy