Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 11

int_0^(pi/2)sin^2x cos^2xdx
By squaring the formula for sine of double angle sin2theta=2sin theta cos theta we get sin^2 2theta=4sin^2theta cos^2theta
1/4int_0^(pi/2)sin^2 2xdx
Make substitution: t=2x=>dx=dt/2. New limits of integration are t_1=2\cdot0=0 and t_2=2cdot pi/2=pi
1/8int_0^pi sin^2 tdt
Use formula for cosine of double angle to obtain sin^2theta=(1-cos2theta)/2
1/16int_0^pi(1-cos2t)dt
Make substitution: u=2t=>dt=(du)/2. New limits of integration are u_1=0 and u_2=2pi.
1/32int_0^(2pi)(1-cos u)du=1/32(u-sin u)|_0^(2pi)=1/32(2pi-0-0+0)=pi/16
The integral is equal to pi/16.