We have the integral #I_(color(red)1 )=intcos^4(2x)*sin^3(2x)dx#
Here, let
#(2x)=t#
#d/dx(2x)=d/dxt#
#2=dt/dx#
Therefore, #dx=dt/2#
Put this in #I_(color(red)1#
#I_(color(red)2 )=1/2intcos^4(t)*sin^3(t)dt#
#I_(color(red)2 )=1/2intcos^4(t)*sin^2(t)*sin(t)dt#
We know #(sin^2t=1-cos^2t)#
Put this in #I_(color(red)2#
#I_(color(red)2 )=1/2intcos^4(t)*(1-cos^2t)*sin(t)dt#
Here, let
#cost=v#
#d/dtcost=d/dtv#
#-sint=(dv)/dt#
Therefore,
#sin(t)dt=-dv#
Put this in #I_(color(red)2#
#I_(color(red)3 )=(-1)/2intv^4*(1-v^2)*dv#
#I_(color(red)3 )=(-1)/2int(v^4-v^6)*dv#
#I_(color(red)3 )=((-1)/2)(v^5)/5-((-1)/2)v^7/7+C#
Now put in the value of #v#.
#I_(color(red)3 )=((-1)/2)(cos^5(t))/5-((-1)/2)cos^7(t)/7+C#
Now put in the value of #t#.
#I_(color(red)3 )=(-cos^5(2x))/10+cos^7(2x)/14+C#
Therefore,
#intcos^4(2x)*sin^3(2x)dx=cos^7(2x)/14-cos^5(2x)/10+C#
