Find the area bounded by the curve y=(sin(x))^2 and the x-axis, between the points x=0 and x=pi/2

First, use the identity cos(2x)=(cos(x))^2-(sin(x))^2 along with the identity (sin(x))^2+(cos(x))^2=1 to obtain the integral of 1/2*(1-cos(2x)) as it is not possible to integrate (sin(x))^2 straight off with a substitution of u=sin(x). Integrating this gives 1/2*(x+2sin(2x)) between x=pi/2 and x=0Evaluating this gives 1/2*(pi/2 +2sin(pi)-0-2sin(0)). Since sin(pi) and sin(0) are both equal to zero, this yields the answer pi/4. Hence the area is pi/4 units^2.

Answered by Thomas L. Maths tutor

4119 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is the chain rule, product rule and quotient rule and when do I use them?


evaluate the integral of lnx


Find the area bounded by the curve x^3-3x^2+2x and the x-axis between x=0 and x=1.


Given the function f(x) = (x^2)sin(x), find f'(x).


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