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integration by substitution - HSN forum

# integration by substitution

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### #1Hev

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Posted 06 November 2006 - 09:24 PM

i really can't get my head around these questions, i managed a few easier ones but this one is just getting me good

using the substituion x= 1 + sin[theta] evaluate

integral of cos[theta] over (1+sin[theta])^3 with respect to theta

which is defined by pie/2 and 0

anyone help me?

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### #2will_789

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Posted 06 November 2006 - 09:40 PM

QUOTE(Hev @ Nov 6 2006, 09:24 PM)

i really can't get my head around these questions, i managed a few easier ones but this one is just getting me good

using the substituion x= 1 + sin[theta] evaluate

integral of cos[theta] over (1+sin[theta])^3 with respect to theta

which is defined by pie/2 and 0

anyone help me?

I can give a bit of advice.

By using the substitution x=1+sin[theta]

therefore u=1+sin[theta]
therefore du = cos[theta]

This helping at all?

### #3Hev

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Posted 06 November 2006 - 09:41 PM

that's as far as i got by myself, then i got really stuck, it's always the next bit i mess up.

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### #4Steve

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Posted 06 November 2006 - 09:53 PM

That's not quite right, it's , so .

Also note that it's easier to change the limits of integration so that they are in terms of x.

Let's do that. The lower limit is in which case and the upper limit is so .

Then this gives:

Note that the whole point of using a substitution is to make the integration easier, notice how the s cancel.

I hope that helps.
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### #5Hev

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Posted 06 November 2006 - 09:59 PM

that's so confussing! i'm really not good at this

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### #6Steve

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Posted 06 November 2006 - 10:01 PM

Where do you lose track of it?

(By the way, different techers will teach this differently.)
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### #7Hev

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Posted 06 November 2006 - 10:03 PM

ok, so when u have the d[theta] that means times by d[theta] so u change it for dx/cos[theta]?

we have two teachers for the advanced course, and one of them i understand but the other i don't follow her style of teaching atall, n it's her that's teaching us this.

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### #8Steve

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Posted 06 November 2006 - 10:08 PM

Technically no, but yes in practice! It's a nice property of the chain rule.
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### #9Hev

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Posted 06 November 2006 - 10:21 PM

ok i've got it now, thanks i pritty much got the most of it, except where u changed the limits? i'm not sure if we actually did those in class, i don't have them in my notes, but i miss out 40 minutes a week due to a double period class with advanced higher biology classes so i might just have missed it out in class. i'll go back and try n learn that better. calculus is not my favourite i must say. but thank you for all your help

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### #10maximus

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Posted 19 September 2007 - 10:04 PM

I'm slightly confused by what the xdx on top of this function does to it, i've tried quite a few ways of doing it and i nearly get the right answer but im missing another expression to the power of something

xdx/the square root of (2+4x)

thanks

### #11Steve

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Posted 20 September 2007 - 07:11 PM

An important thing to remember is that is just another way of writing .

Usually with things like this, you would use the substitution , then take it from there. So you have so .

But when we make the substitution, we want to get rid of the xs if we can. So we can write x in terms of u by rearranging to get . So the integral becomes:

Which is now just like one you could have done in Higher .

Once you have integrated, you might want to have an expression in x rather than in u, so you can use the substitution again.

Does this help at all? (If not, I'll try to help more.)
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### #12maximus

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Posted 21 September 2007 - 04:06 PM

ahhhhh thats great thanks, i did it like that at first, then the person next to me said that i was doing it wrong

thanks alot, its quite simple like that!

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