As I understand it, the substitution rule is:
$$\int f(g(x)).g'(x) \; dx=\int f(u) \; du\text{ where }u=g(x)$$
I had to solve the following:
$$\int \sin^6x \cos^3x dx=\int \sin^6x(1-\sin^2x)\cos x\;dx$$
I understand it this far.
The text explained that the substitution here was
$$g(x) = u = \sin x$$
Which lead to solving the following:
$$\int u^6(1-u^2) \; du$$
I don't understand this. This isn't the form in the substitution rule. If $g(x)=\sin x$ then it would have been
$$f(x)=1-x^2$$
and
$$\int f(g(x))g'(x)dx = \int (1-\sin^2 x)\cos x \;dx$$
I don't know how to include the $\sin^6x$.
Can someone explain how this works?