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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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The function $f(u)$ is $u^6(1-u^2)$. –  André Nicolas Feb 27 '12 at 22:38
    
See the comment by André Nicolas above, and then write what you've got as $u^6 - u^8$ and go on from there. –  Michael Hardy Feb 27 '12 at 23:58
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1 Answer 1

up vote 3 down vote accepted

$\bf Hint:$ 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. –  Korgan Rivera Feb 27 '12 at 22:45
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