Use integration by parts to evaluate: ∫xsin(x) dx.

Since our function is a product of two "mini-functions" of x, we are able to use integration by parts.The trick for this is to correctly set 'u' and 'dv'. 'u' should be labelled as the function which can reduce when differentiated. This means that the function should decrease in power. From our main function, we have both x, and sin(x). If we differentiate sin(x), we get cos(x), which hasn't decreased. However, if we differentiate x, we get 1, which has decreased in power.Using the integration by parts formula: ∫[udv] = uv - ∫[vdu].By setting u=x, and dv=sin(x), we can calculate 'du' and 'v' by differentiating u, and integrating dv respectively.u=x --> du=1, and dv=sin(x) --> v= -cos(x).Substituting this into the integration by parts formula stated above gives:∫[xsin(x)] = x*(-cos(x)) - ∫[(-cos(x)1)],Therefore ∫[xsin(x)] = -xcos(x) - ∫[(-cos(x)],And finally: ∫[xsin(x)] = -xcos(x) - (-sin(x)).This gives us our final answer of:∫[xsin(x)] = -xcos(x)+sin(x) + C, where C is the constant of integration.

Answered by Bailey A. Maths tutor

2903 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has the equation 6x^(3/2) + 5y^2 = 2 (a) By differentiating implicitly, find dy/dx in terms of x and y. (b) Hence, find the gradient of the curve at the point (4, 3).


g(x) = e^(x-1) + x - 6 Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, where x<6


How do I rewrite 2 cos x + 4 sin x as one sin function?


Find the value of 2∫1 (6x+1) / (6x2-7x+2) dx, expressing your answer in the form mln(2) + nln(3), where m and n are integers.


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 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences