How to solve ∫ (sin^4x)(cos^2x) dx ?
1 Answer
Explanation:
intsin^4x*cos^2x dx=int(sin^2x)*(sin^2x*cos^2x)dx∫sin4x⋅cos2xdx=∫(sin2x)⋅(sin2x⋅cos2x)dx
=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx=∫(12)⋅(1−cos2x)⋅(sin2x2)2dx
We have
cos2x=1-2sin^2xcos2x=1−2sin2x
sin^2x=(1-cos2x)/2sin2x=1−cos2x2
and
sin2x=2sinx*cosxsin2x=2sinx⋅cosx
The integral becomes
=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx=∫(12)⋅(1−cos2x)⋅(sin2x2)2dx
=1/8(int(1-cos2x)*sin^2(2x))dx=18(∫(1−cos2x)⋅sin2(2x))dx
=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx=18(∫(sin2(2x)−cos2x⋅sin2(2x))dx
=1/8(int(1/2)*2sin^2(2x)dx-int1/2*2cos2x*sin^2(2x)dx)=18(∫(12)⋅2sin2(2x)dx−∫12⋅2cos2x⋅sin2(2x)dx)
=1/8(int(1/2)*(1-cos(4x))dx-int1/2*sin^2(2x)d(sin2x))=18(∫(12)⋅(1−cos(4x))dx−∫12⋅sin2(2x)d(sin2x))
=1/16(int(1-cos(4x))dx-intsin^2(2x)d(sin2x))=116(∫(1−cos(4x))dx−∫sin2(2x)d(sin2x))
d(sin2x)/dx=2cos2xdsin2xdx=2cos2x
d(sin2x)=2cos2x*dxd(sin2x)=2cos2x⋅dx