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

1 Answer
Jul 28, 2014

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

Explanation,

=intsin^2(2t)dt

Using trigonometric identity, cos2t=1-2sin^2t

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

=int(1-cos4t)/2dt

=1/2int1dt-1/2intcos4tdt

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

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