Let, I=intcos^4t*sin^2tdt.
Observe that, cos and sin both have even Power. So, we have
to convert them into Multiple Angles, using the Identities,
(1): cos^2theta=(1+cos2theta)/2, (2): sin^2theta=(1-cos2theta)/2,
(3): 2cosalphacosbeta=cos(alpha+beta)+cos(alpha-beta).
Now, cos^4t*sin^2t=1/4(4cos^2t*sin^2t)(cos^2t),
=1/4(sin^2(2t))(cos^2t),
=1/4{1/2(1-cos4t)}{1/2(1+cos2t)},
=1/16(1-cos4t+cos2t-cos4tcos2t),
=1/32{2-2cos4t+2cos2t-2cos4tcos2t},
=1/32{2-2cos4t+2cos2t-(cos6t+cos2t)},
=1/32{2-cos6t-2cos4t+3cos2t}.
Therefore, I=int1/32{2-cos6t-2cos4t+3cos2t}dt,
=1/32(2t-1/6sin6t-2*1/4sin4t+3*1/2sin2t),
rArr I=1/192(12t-sin6t-3sin4t+9sin2t)+C.
Enjoy Maths.!
