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?