Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 10

You need to use the fundamental trigonometric formula sin^2 x = 1 - cos^2 x:
int sin^2 t*cos^4 t dt = int (1 - cos^2 t)*cos^4 t dt
int (1 - cos^2 t)*cos^4 t dt = int cos^4 t dt - int cos^6 t dt
You should use the following formula:
cos^2 t = (1 + cos 2t)/2 => cos^4 t = ((1 + cos 2t)^2)/4
cos^6 t = ((1 + cos 2t)^3)/8
int cos^4 t dt= (1/4) int ((1 + cos 2t)^2) dt

int cos^4 t dt= (1/4) int dt + (1/4) int 2cos 2t dt + (1/4) cos^2 2t dt

int cos^4 t dt= (1/4) t + (1/4) sin 2t + (1/4)int (1 + cos 4t)/2dt
int cos^4 t dt= (1/4) t + (1/4) sin 2t + (1/8)( t + (sin 4t)/4) + c
You need to solve int cos^6 t dt such that:
int cos^6 t dt = int ((1 + cos 2t)^3)/8 dt
int cos^6 t dt = (1/8)int dt + (1/8) int cos^3 2t dt + (3/8) int cos^2 2t dt +(3/8) int cos 2t dt
(1/8) int cos^3 2t dt = (1/8) int cos^2 2t *cos 2t dt
int cos^2 2t *cos 2t dt = int (1 - sin^2 2t) *cos 2t dt
sin 2t = u => 2cos 2t dt = du
int (1 - sin^2 2t) *cos 2t dt = int (1 - u^2) *(du)/2
int (1 - u^2) *(du)/2 = u/2 - u^3/6
int (1 - sin^2 2t) *cos 2t dt = (sin 2t)/2 - (sin^3 2t)/6
int cos^6 t dt = (1/8)t + (1/8)((sin 2t)/2 - (sin^3 2t)/6) + (3/16) sin 2t + (3/8) int cos^2 2t dt
int cos^6 t dt = (1/8)t + (1/8)((sin 2t)/2 - (sin^3 2t)/6) + (3/16) sin 2t + (3/8) int (1 + cos 4t)/2 dt
int cos^6 t dt = (1/8)t + (1/8)((sin 2t)/2 - (sin^3 2t)/6) + (3/16) sin 2t + (3/8) (t + (sin 4t)/4)
Hence, the result of integration is:
int sin^2 t*cos^4 t dt = (1/4)pi + (1/4) sin 2pi + (1/8)(pi + (sin 8pi)/4)- (1/8)(pi - (1/8)((sin 2pi)/2 - (sin^3 4pi)/6)- (3/16) sin4pi- (3/8) (pi/2 + (sin 8pi)/4)
int sin^2 t*cos^4 t dt = int sin^2 t*cos^4 t dt = pi/4 - (3pi)/16 = pi/16