Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 9

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