How to Use the Half-Angle Formula for (cosx)^2?

[nameVoid]

y=sin2x bounded by x=0,x=pi,y=0 , revolved around the x-axis

cross section A=pi(sin2x)^2

taking u = sinx ; du=cosxdx

im unclear on how to proceed in this case where du needs to satisfy (cosx)^2
the problem hints to use a half angle formula

 

[slider142]

Try the identity cos(2x) = 1 - 2sin²(x). It can be rearranged to be a half-angle formula.

 

[nameVoid]

still i am unclear on how to proceed here

 

[slider142]

still i am unclear on how to proceed here

Try using the Pythagorean identity to get sin²(x) - sin⁴(x), then apply the half angle formula to get cosines (twice for the second term) that are not squared.

 

[Dick]

slider142 is trying to say that since sin^2(x)=(1-cos(2x))/2, sin^(2x)=(1-cos(4x))/2. That's pretty easy to integrate.

 

[nameVoid]

perfect