`int_0^pi cos^4(2t) dt` Evaluate the integral

Let us call our integral `I.`

`I=int_0^pi cos^4 2t dt=int_0^pi(cos^2 2t)^2dt`

Use formula for cosine of double angle: `cos^2 theta=(1+cos2theta)/2`

`int_0^pi((1+cos4t)/2)^2dt=1/4int_0^pi(1+2cos4t+cos^2 4t)dt=`

`1/4int_0^pi dt+1/2int_0^pi cos4tdt+1/4int_0^picos^2 4tdt`

Let us denote the above three integrals by `I_1,I_2` and `I_3` respectively i.e.

`I=1/4I_1+1/2I_2+1/4I_3`

`I_1=t|_0^pi=pi`

To solve `I_2` we make substitution `u=4t=>(du)/4=dt` with new limits of integration `u_1=4cdot0=0` and `u_2=4cdotpi=4pi.`

`I_2=1/4int_0^(4pi)cos u du=1/4sin u|_0^(4pi)=0`

To calculate `I_3` we use formula for cosine of double angle once again.

`I_3=int_0^pi(1+cos8t)/2 dt`

Make substitution `u=8t=>(du)/8=dt` with new limits of integration `u_1=8cdot0=0` and `u_2=8cdotpi=8pi`

`1/16int_0^(8pi)(1+cos u)du=1/16(u+sin u)|_0^(8pi)=1/16(8pi+0-0-0)=pi/2`

Now that we have calculated the three integrals we can return to calculate `I.`

`I=1/4cdot pi+1/2cdot0+1/4cdotpi/2=pi/4+pi/8=(3pi)/8`