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### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1265283 times) Tweet Share

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#### schoolstudent115

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9450 on: April 21, 2019, 07:19:28 pm »
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Hi I can't figure out how to do part b from both questions linked. For question 1 I can't figure out how to use recognition to rearrange the equation (which I'm pretty sure is the method they want you to use) and I only got so far as rearranging for the integral of xcos^n(x) from the original equation. Question 2 they most likely expect you to use recognition too. I tried to change all the x's in part a) ii into (pi/2 - x) and then isolate the xcos(x) integral, but got (pi-x)sin(x)-cos(x) as a final answer which is wrong too (I know using integration by parts would make things 20 times less complicated but I need those working marks if I get the wrong answer). As always help is appreciated.
Question 1: From what you said, you did part (a),  you should have gotten: $\cos^n(x)-(n-1)\sin^2(x)\cos^{n-2}(x)$ (using the chain rule and product rule).

So $\frac{d}{dx}[\sin(x)\cos^{n-1}(x)]=\cos^n(x)-(n-1)\sin^2(x)\cos^{n-2}(x)$. Rearranging for cos^n(x)

$\frac{d}{dx}[\sin(x)\cos^{n-1}(x)]+(n-1)\sin^2(x)\cos^{n-2}(x)=\cos^n(x)$
Integrating both sides (and flipping the equation): $\int{\cos^n(x)}\mathrm{d}x=\int{\frac{d}{dx}[\sin(x)\cos^{n-1}(x)]}\mathrm{d}x+(n-1)\int{\sin^2(x)\cos^{n-2}(x)}\mathrm{d}x$

Let $I=\int{\cos^n(x)}\mathrm{d}x$

Notice that the integrand on the RHS of the equation from before is just $\sin^2(x)\cos^{n-2}(x)=(1-\cos^2(x))*\cos^{n-2}(x)=\cos^{n-2}(x)-\cos^n(x)$
So $\int{\sin^2(x)\cos^{n-2}(x)}\mathrm{d}x=\int{\cos^{n-2}(x)-\cos^n(x)}\mathrm{d}x=\int{\cos^{n-2}(x)}\mathrm{d}x-\int{cos^n(x)}\mathrm{d}x=\int{\cos^{n-2}(x)}\mathrm{d}x-I$

So substituting the previous result back in:
$I=\sin(x)\cos^{n-1}(x)+(n-1)(\int{\cos^{n-2}(x)}\mathrm{d}x-I)=\sin(x)\cos^{n-1}(x)+(n-1)(\int{\cos^{n-2}(x)}\mathrm{d}x)-I(n-1)$
Adding $I(n-1)$ to both sides $\implies I+I(n-1)=\sin(x)\cos^{n-1}(x)+(n-1)(\int{\cos^{n-2}(x)}\mathrm{d}x)$

The left hand side is just $nI$, So divide through by $n$ to get the final value of the integral:
$I=\frac{1}{n}\sin(x)\cos^{n-1}(x)+\frac{n-1}{n}\int{\cos^{n-2}(x)}\mathrm{d}x$

As for the second question you had, you may have to use a similar technique. If you need further help feel free to reply.

« Last Edit: April 22, 2019, 11:26:50 am by schoolstudent115 »
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#### undefined

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9451 on: April 22, 2019, 09:04:03 pm »
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#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9452 on: April 23, 2019, 04:32:36 pm »
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Hi I have a question about linear substitution in relation this question:

∫x^2 √(x-1)

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9453 on: April 23, 2019, 05:12:37 pm »
0
Hi I have a question about linear substitution in relation this question:

∫x^2 √(x-1)

The substitution  $u=x-1$  should give you progress

Spoiler
Using  $u=x-1$,  we have  $x^2=(u+1)^2$  and  $\dfrac{du}{dx}=1$,  so
$\int x^2\sqrt{x-1}\,\text{d}x=\int (u+1)^2\sqrt{u}\,\text{d}u=\int\left(u^{5/2}+2u^{3/2}+u^{1/2}\right)\text{d}u,$ which can be integrated quite easily.
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#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9454 on: April 23, 2019, 05:24:09 pm »
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Thankyou, I did use this approach originally, however after integrating I attained solution of:

2/7(x-1)^(7/2)+2/5(x-1)^(5/2)+3/2(x-1)^(3/2)+c

When the actual solution is:

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9455 on: April 23, 2019, 05:37:15 pm »
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Thankyou, I did use this approach originally, however after integrating I attained solution of:

2/7(x-1)^(7/2)+2/5(x-1)^(5/2)+3/2(x-1)^(3/2)+c

When the actual solution is:

The solution provided just does some really unnecessary factoring by taking out a  $2/105$  and  $(x-1)^{3/2}$  from every term.

Also, just be a little careful when applying the power rule. Correct answer in expanded form is: $\int x^2\sqrt{x-1}\,\text{d}x=\frac{2}{7}(x-1)^{7/2}+\frac{4}{5}(x-1)^{5/2}+\frac{2}{3}(x-1)^{3/2}+c,\quad c\in\mathbb{R}$
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#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9456 on: April 23, 2019, 05:56:18 pm »
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Right, I appreciate the assistance. I'll make sure to look out for those errors. Though I'm curious, is it necessary to state that c is part of R (soz can't remember the name for the symbol) with the final solution?

And thank you for the help overall, my first experience using the forums was a positive one.

#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9457 on: April 23, 2019, 06:02:11 pm »
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Right, I appreciate the assistance. I'll make sure to look out for those errors. Though I'm curious, is it necessary to state that c is part of R (soz can't remember the name for the symbol) with the final solution?

And thank you for the help overall, my first experience using the forums was a positive one.

No, you don't need to specify that $c$ is a real constant. It's pretty clear in the context of integration what $c$ is. However, in a situation where you have to deal with many variables, it might be nice to be precise and label what everything is for the sake of clarity. The only reason I specified it is because I've gotten into the good habit of doing so.

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#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9458 on: April 25, 2019, 10:52:33 am »
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Hi again, another question in relation to inverse circular functions:

How would the the implied domain and range be affected by squaring x :

y=cos^(-1) (x^2)

I can understand that the domain of [-1,1] will not be affected, however what would be the calculations for deriving the range?

#### RuiAce

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9459 on: April 25, 2019, 11:21:14 am »
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Hi again, another question in relation to inverse circular functions:

How would the the implied domain and range be affected by squaring x :

y=cos^(-1) (x^2)

I can understand that the domain of [-1,1] will not be affected, however what would be the calculations for deriving the range?

At this point, now we note that the range of $h(x)=x^2$ when restricted to domain $[-1,1]$ is $[0,1]$.

The range you're after will now coincide with the range of $g(x)=\cos^{-1}x$, but with domain of $g$ restricted to $[0,1]$. (i.e. Restrict the domain of $g$ to the range of $h$.) A sketch is then enough to help us deduce that the range will be $\left[0, \frac\pi2\right]$.
« Last Edit: April 25, 2019, 11:22:45 am by RuiAce »

#### persistent_insomniac

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9460 on: April 25, 2019, 11:34:38 am »
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How do you integrate 1/9+4x^2 ? I get it is tan^-1(2x/3) + c but how do you get the 1/6 at the front?

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9461 on: April 25, 2019, 11:45:26 am »
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How do you integrate 1/9+4x^2 ? I get it is tan^-1(2x/3) + c but how do you get the 1/6 at the front?

Recall that:

$\int \frac{a}{a^2+x^2}dx=\textrm{arctan}(\frac{x}{a})$

Hence:

$\int \frac{1}{9+4x^2}dx=\int \frac{1}{4(\frac{9}{4}+x^2)}dx=\frac{1}{4}\int \frac{1}{(\frac{3^2}{2^2}+x^2)}dx=\frac{1}{4} \times \frac{2}{3} \int \frac{\frac{3}{2}}{(\frac{3^2}{2^2}+x^2)}dx = \frac{1}{6}\textrm{arctan}(\frac{2x}{3})+c$

Since in this case a = 3/2 and we have a factor of 1/6 out the front of the integral.

#### Flight112

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9462 on: April 25, 2019, 12:37:29 pm »
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At this point, now we note that the range of $h(x)=x^2$ when restricted to domain $[-1,1]$ is $[0,1]$.

The range you're after will now coincide with the range of $g(x)=\cos^{-1}x$, but with domain of $g$ restricted to $[0,1]$. (i.e. Restrict the domain of $g$ to the range of $h$.) A sketch is then enough to help us deduce that the range will be $\left[0, \frac\pi2\right]$.

Ok, but how is putting the domain with the restrictions [-1,1] into the h(x)=x^2 equate to the restriction [0,1]?

#### RuiAce

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9463 on: April 25, 2019, 12:43:02 pm »
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Ok, but how is putting the domain with the restrictions [-1,1] into the h(x)=x^2 equate to the restriction [0,1]?
If you sketch $y=x^2$ with domain restriction $[-1,1]$, don't the values for $y$ become restricted to being between 0 and 1?

#### Flight112

If you sketch $y=x^2$ with domain restriction $[-1,1]$, don't the values for $y$ become restricted to being between 0 and 1?