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
