How would the integral ∫x^2sin2xdx be solved using integration by parts?

The general formula for integration by parts is given as
∫u(dv/dx)dx = uv - ∫v(du/dx)dx
given that the equation to be solved is ∫x2sin2xdx, the values for u, v, du/dx and dv/dx can be assigned as
u = x2 du/dx = 2x
v = (-1/2)cos2x dv/dx = sin2x
These values can then be plugged into the general formula to solve the integral
∫x2sin2xdx = (-1/2)x2cos2x + ∫xcos2x dx
the second integral is then solved. In this instance, integration by parts has to be used a second time. The new values for u, v, du/dx and dv/dx can be assigned as
u = x du/dx = 1
v = (1/2)sin2x dv/dx = cos2x
Therefore the integral can be solved as
∫x2sin2xdx = (-1/2)x2sin2x + (1/2)xsin2x - ∫(1/2)sin2x dx
= (-1/2)x2sin2x + (1/2)xsin2x + (1/4)cos2x + C

Answered by Samuel N. Maths tutor

9854 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you differentiate a function comprised of two functions multiplied together?


The curve C has equation y = x^3 - 2x^2 - x + 9, x > 0. The point P has coordinates (2, 7). Show that P lies on C.


Express [1+4(square root)7] /[ 5+ 2(square root)7] in the form m + n (square root)7 , where m and n are integers.


A curve has the equation y=x^3+2x+15. Find dy/dx.


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