We know that,
(1)sin^2theta=(1-cos2theta)/2(1)sin2θ=1−cos2θ2
(2)cos^2 2theta=(1+cos4theta)/2(2)cos22θ=1+cos4θ2
(3)intcosAxdx=1/AsinAx+c(3)∫cosAxdx=1AsinAx+c
Here,
I=∫ sin^4 x dxI=∫sin4xdx
=int(sin^2x)^2dx..toApply(1)=∫(sin2x)2dx..→Apply(1)
=int((1-cos2x)/2)^2dx=∫(1−cos2x2)2dx
=1/4int(1-2cos2x+cos^2 2x)...toApply(2)
=1/4int(1-2cos2x+(1+cos4x)/2)dx
=1/8int(2-4cos2x+1+cos4x)dx
=1/8int(3-4cos2x+cos4x)dx...toApply(3)
=1/8[3x-(4sin2x)/2+(sin4x)/4]+c
=1/32[12x-8sin2x+sin4x]+c