`int_0^(pi/2) sin^2(x) cos^2(x) dx` Evaluate the integral

`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.`