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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?

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    See the comment by André Nicolas above, and then write what you've got as $u^6 - u^8$ and go on from there.2012-02-27

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$\bf Hint:$ Consider $f(x)=x^6(1-x^2)$

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    lol it's laughably obvious isn't it! Why do I bother asking these questions. Ok thanks. I didn't realise I could do that.2012-02-27