How do I evaluate the indefinite integral intsin^2(2t)dtsin2(2t)dt ?

1 Answer
Jul 28, 2014

=1/2t-1/8sin4t+c=12t18sin4t+c, where cc is a constant

Explanation,

=intsin^2(2t)dt=sin2(2t)dt

Using trigonometric identity, cos2t=1-2sin^2tcos2t=12sin2t

sin^2t=(1-cos2t)/2sin2t=1cos2t2, inserting this value of sin^2tsin2t in integral, we get

=int(1-cos4t)/2dt=1cos4t2dt

=1/2int1dt-1/2intcos4tdt=121dt12cos4tdt

=1/2t-1/2(sin4t)/4+c=12t12sin4t4+c, where cc is a constant

=1/2t-1/8sin4t+c=12t18sin4t+c, where cc is a constant