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

5401 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve 4log₂(2)+log₂(x)=3


Use chain rule and implicit differentiation to find dy/dx for y^3 = 1 + 3*x^2, then show that they are equal


Show that r^2(r + 1)^2 - r^2(r - 1)^2 ≡ 4r^3.


Find the exact gradient of the curve y=ln(1-cos2x) at the point with x-coordinate π/6


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:

© 2026 by IXL Learning