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

Related Maths A Level answers

All answers ▸

A new sports car accelerates using rockets at 5m/s for 30 seconds from some traffic lights and then decelerate for 45 seconds to a stop.


How to find and classify stationary points (maximum point, minimum point or turning points) of curve.


Express as a simple logarithm 2ln6 - ln3 .


Differentiate (x^0.5)ln(x) with respect to x.