What is the integral of x sin(x) dx?

Find the following integral: ∫ x sin(x) dx

This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.

Step 1) Recall that if you have an integral of the form:
∫ u(dv/dx) dx

Then it can be written as:
uv – ∫ v(du/dx) dx

We need to decide which part we will differentiate (as in, which part is u), and which part we will integrate (as in, which part is dv/dx).

Step 2) We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1. From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x)).

So, let:
u = x, which implies du/dx = 1

And let:
dv/dx = sin(x). Integrating this to get v gives v = –cos(x)

Step 3) So, our integral is now of the form required for integration by parts.
∫ x sin(x) dx 
= ∫ u(dv/dx) dx
= uv –  ∫ v(du/dx) dx
= –x cos(x) – ∫ –cos(x)*1 dx
= –x cos(x) – ∫ –cos(x) dx
= –x cos(x) + ∫ cos(x) dx

The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x).

Step 4) Finally, as with all integration without limits, there must be a constant added, which I'll call c. So the final answer is:

∫ x sin(x) dx = –x cos(x) + sin(x) + c

Answered by Shaun F. Maths tutor

303704 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I find the maxima and minima of f(x) = e^(x^2)?


A curve is defined by the parametric equations x=t^2/2 +1 and y=4/t -1. Find the gradient of the curve at t=2 and an equation for the curve in terms of just x and y.


f (x) = (x^2 + 4)(x^2 + 8x + 25). Find the roots of f (x) = 0


For sketching the graph of the modulus of f(x) (in graph transformations), why do we reflect in the x-axis anything that is below it?


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
Cookie Preferences