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.

TL
Answered by Thomas L. Maths tutor

4739 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

If y = exp(x^2), find dy/dx


Do the circles with equations x^2 -2x + y^2 - 2y=7 and x^2 -10x + y^2 -8y=-37 touch and if so, in what way (tangent to each other? two point of intersection?)


A curve has equations: x=2sin(t) and y=1-cos(2t). Find dy/dx at the point where t=pi/6


What does dy/dx represent?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning