Solve this integral without reduction formulae?

Rearrange it as sin^3(ax) * cos(ax) * [1-sin^2(ax)] dx

Now u=sin(ax); du=acos(ax) dx and du/a =cos(ax) dx

(1/a) INTEGRAL[u^3(1-u^2) du]

=(1/a)INTEGRAL[u^3-u^5 du]

=(1/a)(u^4/4 - u^6/6) + C

= [sin^4(ax)/(4a)] + [sin^6(ax)/(6a)] + C

integral of sin^3(ax)cos^3(ax)

= sin(ax)sin^2(ax)cos^3(ax)

= sin(ax)(1-cos^2(ax))cos^3(ax)

u = cos(ax)

du = a dx

= 1/a integral of (1 - u^2)*u^3 du

= 1/a(1/4*u^4 - 1/6*u^6) + C

Now plug back in cos(ax)

= 1/a(1/4*cos^4(ax) - 1/6*cos^6(ax)) + C

1) Applying the identity, sin(2A) = 2sin(A)cos(A),

sin³(ax)cos³(ax) = (1/8){2sin(ax)cos(ax)}³ = (1/8)sin³(2ax)

2) Next from another identity, sin(3A) = 3sin(A) - 4sin³A, sin³A = (1/4){3sin(A) - sin(3A)}

Applying this, the above further changes to: (1/32){3sin(2ax) - sin(6ax)}

3) Integrating this, [sin(ax) = -{cos(ax)}/a]

= (1/32){-(3/2a)cos(2ax) + (1/6a)cos(6ax)} + C

Simplifying further, answer = (1/192a)*{cos(6ax) - 9cos(2ax)} + C

int(sin^3*a*x*cos^3*(a*x), x) = (a^2/3)*sin^3*x^3*cos^3