Indefinite Integral Primitives? (1 Viewer)

Finx · Feb 13, 2009

Find ∫x³(x^4 - 1)^8 dx

I know the rule for finding ∫(ax+b)^n etc, but the x³ really throws me off. I could expand, but that'd be extremely tedious and most likely the wrong approach for the question.

Thanks in advance!

tommykins · Feb 13, 2009

Finx · Feb 13, 2009

I love you.

Trebla · Feb 13, 2009

Remember the chain rule in differentiation:
$\dfrac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$
Therefore we can say that:
$\displaystyle\int{f'(g(x))g'(x)}\,dx = f(g(x))+c$
Here f(x) = x⁹ giving f'(x) = 9x⁸ and g(x) = x⁴-1 giving g'(x) = 4x³
$\displaystyle\int{x^3(x^4-1)^8}\,dx = \dfrac{1}{36}\displaystyle\int{4x^39(x^4-1)^8}\,dx = \dfrac{1}{36}(x^4-1)^9 + c$

tommykins · Feb 13, 2009

Trebla said:
Remember the chain rule in differentiation:
$\dfrac{d}{dx}(f(g(x)))=f'(g(x))g'(x)$
Therefore we can say that:
$\displaystyle\int{f'(g(x))g'(x)}\,dx = f(g(x))+c$
Here f(x) = x⁹ giving f'(x) = 9x⁸ and g(x) = x⁴-1 giving g'(x) = 4x³
$\displaystyle\int{x^3(x^4-1)^8}\,dx = \dfrac{1}{4}\displaystyle\int{4x^3(x^4-1)^8}\,dx = \dfrac{1}{36}(x^4-1)^9 + c$

I used to do that during that during HSC but there was something somewhere that a teacher said you can't do that cause technically it's wrong.

Forgot what it was though.

Trebla · Feb 13, 2009

There is nothing technically wrong with that. It's a method even encouraged in first year uni integration, not to mention quicker if you're good at recognising the form...If anything your substitution has a small technical issue because you have x's and u's in the same integrand when you're not supposed to...lol

Also, I forgot to mention that since 2 unit students do not learn integration by substitution, they are expected to know how to use the reverse chain rule instead lol

Finx · Feb 13, 2009

Fight fight fight fight fight

(Thanks for your input too Trebla =D)

tommykins · Feb 13, 2009

Trebla said:
There is nothing technically wrong with that. It's a method even encouraged in first year uni integration, not to mention quicker if you're good at recognising the form...If anything your substitution has a small technical issue because you have x's and u's in the same integrand when you're not supposed to...lol

Also, I forgot to mention that since 2 unit students do not learn integration by substitution, they are expected to know how to use the reverse chain rule instead lol

Yeah fair enough hey

Sarah182 · Feb 14, 2009

Hmmm Intergration by substitution is a 3 unit topic??
You shouldn't get any intergration questions this hard in the HSC, cause really you shouldn't have learned how to do them?

x.Exhaust.x · Feb 14, 2009

Sarah182 said:
Hmmm Intergration by substitution is a 3 unit topic??
You shouldn't get any intergration questions this hard in the HSC, cause really you shouldn't have learned how to do them?

LOL this is hard? It's simple integration by substitution.

lolokay · Feb 14, 2009

^ the point is, in 2 unit you apparently don't do substitution at all

Aerath · Feb 14, 2009

Yeah, I don't think I needed substitution in 2U at all.

arjungamer123 · Feb 14, 2009

The best way to do it is to pretend it's the actual derivative outside the brackets, then multiply or divide on the outside to get it back to the actual variable given, e.g:
$\int x(x^{2}+1)^{2}= \frac{1}{2}\int 2x(x^{2} + 1)^{2} = \left [\frac{(x^{2}+1)^{3} }{6}\right ]$

x.Exhaust.x · Feb 14, 2009

lolokay said:
^ the point is, in 2 unit you apparently don't do substitution at all

Ah, I thought this was the 3U forum. Sorry.

Sarah182 · Feb 15, 2009

x.Exhaust.x said:
LOL this is hard? It's simple integration by substitution.

No it's not hard, it's just not part of their syllabus?
So technically 2 unit won't have to do any integration THAT hard.

Indefinite Integral Primitives? (1 Viewer)

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