cos(2x)=1-2sin^2(x), sin^2(x)=(1-cos(2x))/2; cos(4x)=2cos^2(2x)-1, cos^2(2x)=(cos(4x)+1)/2.
sin^6(x)=(sin^2(x))^3=(1-cos(2x))^3/8=(1-3cos(2x)+3cos^2(2x)-cos^3(2x))/8.
∫sin^6(x)dx=∫(1-3cos(2x)+3cos^2(2x)-cos^3(2x))dx/8=x/8-(3/16)sin(2x)+(3/64)sin(4x)+3x/16-∫cos^3(2x)dx/8.
[3cos^2(2x)=(3/2)(cos(4x)+1), the integral of which is (3/2)(sin(4x)/4+x)=(3/8)sin(4x)+3x/2.]
∫cos^3(2x)dx=∫cos^2(2x)cos(2x)dx. Integrate by parts: dv=cos(2x)dx, v=sin(2x)/2; u=cos^2(2x), du=-4cos(2x)sin(2x)dx. ∫cos^3(2x)dx=cos^2(2x)sin(2x)/2+2∫sin^2(2x)cos(2x)dx. [d(uv)/dx=vdu/dx+udv/dx; so udv/dx=d(uv)/dx-vdu/dx; therefore, ∫udv=uv-∫vdu, hence integrating by parts using u and v.]
Let p=sin(2x), dp=2cos(2x)dx so 2∫sin^2(2x)cos(2x)dx=∫p^2dp=p^3/3=sin^3(2x)/3.
So (1/8)∫cos^3(2x)dx=cos^2(2x)sin(2x)/16+sin^3(2x)/24.
∫sin^6(x)dx=x/8-(3/16)sin(2x)+(3/64)sin(4x)+3x/16-cos^2(2x)sin(2x)/16-sin^3(2x)/24+k=
5x/16-(3/16)sin(2x)+(3/64)sin(4x)-cos^2(2x)sin(2x)/16-sin^3(2x)/24+k, where k is constant of integration.
This further reduces to:
5x/16-(3/16)sin(2x)+(3/64)sin(4x)-(1-sin^2(2x))sin(2x)/16-sin^3(2x)/24+k=
5x/16-(3/16)sin(2x)+(3/64)sin(4x)-sin(2x)/16+sin^3(2x)/16-sin^3(2x)/24+k.
∫sin^6(x)dx=5x/16-(1/4)sin(2x)+(3/64)sin(4x)+(1/48)sin^3(2x)+k, where k is constant of integration.
By carrying out the conversion to single angles below, I can confirm that this solution using multiple angles is correct.
The multiple angles can be further reduced to single angles, although the question hasn't so requested:
sin(2x)=2sin(x)cos(x);
sin(4x)=2sin(2x)cos(2x)=2(2sin(x)cos(x))(1-2sin^2(x))=4sin(x)cos(x)-8sin^3(x)cos(x);
sin^3(2x)=(2sin(x)cos(x))^3=8sin^3(x)cos(x)(1-sin^2(x))=8sin^3(x)cos(x)-8sin^5(x)cos(x).
Applying these:
∫sin^6(x)dx=5x/16-sin(x)cos(x)/2+(3/16)(sin(x)cos(x)-2sin^3(x)cos(x))+(1/6)(sin^3(x)cos(x)-sin^5(x)cos(x))+k;
∫sin^6(x)dx=5x/16-(5/16)sin(x)cos(x)-(5/24)sin^3(x)cos(x)-(1/6)sin^5(x)cos(x)+k.
Using differentiation, I can confirm that this answer is correct.