How do I evaluate the indefinite integral intsin^2(2t)dt∫sin2(2t)dt ?
1 Answer
Jul 28, 2014
=1/2t-1/8sin4t+c=12t−18sin4t+c , wherecc is a constant
Explanation,
=intsin^2(2t)dt=∫sin2(2t)dt Using trigonometric identity,
cos2t=1-2sin^2tcos2t=1−2sin2t
sin^2t=(1-cos2t)/2sin2t=1−cos2t2 , inserting this value ofsin^2tsin2t in integral, we get
=int(1-cos4t)/2dt=∫1−cos4t2dt
=1/2int1dt-1/2intcos4tdt=12∫1dt−12∫cos4tdt
=1/2t-1/2(sin4t)/4+c=12t−12sin4t4+c , wherecc is a constant
=1/2t-1/8sin4t+c=12t−18sin4t+c , wherecc is a constant