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integration by substitution - HSN forum

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integration by substitution


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#1 Hev

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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 sad.gif

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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#2 will_789

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Posted 06 November 2006 - 09:40 PM

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

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

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?

#3 Hev

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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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#4 Steve

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Posted 06 November 2006 - 09:53 PM

That's not quite right, it's \frac{dx}{d\theta} = \cos\theta, so d\theta = \frac{dx}{\cos\theta}.

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 \theta = 0 in which case x=1+\sin0=1 and the upper limit is \theta = \frac\pi2 so x=1+\sin\frac\pi2=2.

Then this gives:



\begin{align*}\int^{\theta=\frac\pi2}_{\theta=0}\frac{\cos\theta}{(1+\sin\theta)^3}\,d\theta &= \int_{x=1}^{x=2}\frac{\cos\theta}{x^3}\frac{dx}{\cos\theta}\qquad\text{replacing $d\theta$ with $\frac{dx}{\cos\theta}$}\\
&=\int_{x=1}^{x=2} x^{-3}\,dx\end{align*}

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

I hope that helps.
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#5 Hev

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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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#6 Steve

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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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#7 Hev

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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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#8 Steve

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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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#9 Hev

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Posted 06 November 2006 - 10:21 PM

ok i've got it now, thanks smile.gif 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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#10 maximus

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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 smile.gif

#11 Steve

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Posted 20 September 2007 - 07:11 PM

An important thing to remember is that \displaystyle\int{\frac{xdx}{\sqrt{2+4x}} is just another way of writing \displaystyle\int{\frac{x}{\sqrt{2+4x}}\;dx.

Usually with things like this, you would use the substitution u=\text{the thing under the square root} = 2+4x, then take it from there. So you have \frac{du}{dx}=4 so dx=\frac{du}{4}.

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 u=2+4x to get x=\frac{u-2}{4}. So the integral becomes:



\begin{align*}&\int{\frac{x}{\sqrt{2+4x}}\;dx}\\
&= \int{\frac{\frac{u-2}{4}}{\sqrt{u}}\;\frac{du}{4}}\quad\text{making all the substitutions}\\
&= \frac{1}{16}\int\frac{u-2}{\sqrt{u}}\;du\\
&= \frac{1}{16}\int\left(\frac{u}{\sqrt u}-\frac{2}{\sqrt u}\right)\;du\\
&= \frac{1}{16}\int\left(u^{\frac12}-2u^{-\frac12}\right)\;du\\
\end{align*}

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

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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#12 maximus

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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 rolleyes.gif

thanks alot, its quite simple like that!






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