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

4961 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do i solve two linear simultaneous equations 2x+y=7 & 3x-y=8 ?


A curve has equation y = f(x) and passes through the point (4, 22). Given that f'(x) = 3x^2 - 3x^(1/2) - 7, use integration to find f(x), giving each term in its simplest form


Given y = 3x^(1/2) - 6x + 4, x > 0. 1) Find the integral of y with respect to x, simplifying each term. 2) Differentiate the equation for y with respect to x.


If f(x) = (3x-2) / x-5 x>6, find a.) ff(8) b.) the range of f(x) c.) f^-1(x) and state its range.


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