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Step ii 2007 q6

I'm not sure if math questions for STEP are meant to be posted in separate posts. So I apologise if I haven't gotten it right. Feel free to merge this post.

Actual question:


Unparseable latex formula:

\text{Differentiate} \ln(x + \sqrt{3 + x^2}) \text{and} x\sqrtsqrt{3 + x^2}

.
Hence find3+x2dx \text{Hence find} \int \sqrt{3 + x^2} dx .

I'm asking about chewwy's solution to Q6 in STEP II 2007:

http://www.thestudentroom.co.uk/attachment.php?attachmentid=42044&d=1183161671

3+x2dx \int \sqrt{3 + x^2} dx

=3+x2+x23+x2x23+x2dx = \displaystyle \int \sqrt{3 + x^2} + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} dx

First Question --- How would you see this trick --- to add x23+x2x23+x2 \displaystyle \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} to the desired integral?

I know the question asks us to differentiate the functions but I still don't see this trick.

Then use ddx3+x2=... \frac{d}{dx}\sqrt{3 + x^2} = ... to get

3+x2dx+x23+x2x23+x2 \int \sqrt{3 + x^2} dx + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}}

Unparseable latex formula:

\displaystyle = x\sqrt{3 + x^2}- \left \int \frac{x^2}{\sqrt{3 + x^2}} \right



Second Question --- I know how to integrate x23+x2 -\int \frac{x^2}{\sqrt{3 + x^2}} by using the substitution x=atan(u) x = atan(u) . But the STEP solution says that no calculus work is needed for this part --- the integration of 3+x2dx \int \sqrt{3 + x^2} dx . I don't see the derivatives in the first part help me to get x23+x2 \displaystyle -\int \frac{x^2}{\sqrt{3 + x^2}}?

Thanks a lot---
(edited 11 years ago)
Original post by adrienne_om
I'm not sure if math questions for STEP are meant to be posted in separate posts. So I apologise if I haven't gotten it right. Feel free to merge this post.

Actual question:


Unparseable latex formula:

\text{Differentiate} \ln(x + \sqrt{3 + x^2}) \text{and} x\sqrtsqrt{3 + x^2}

.
Hence find3+x2dx \text{Hence find} \int \sqrt{3 + x^2} dx .

I'm asking about chewwy's solution to Q6 in STEP II 2007:

http://www.thestudentroom.co.uk/attachment.php?attachmentid=42044&d=1183161671

3+x2dx \int \sqrt{3 + x^2} dx

=3+x2+x23+x2x23+x2dx = \displaystyle \int \sqrt{3 + x^2} + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} dx

First Question --- How would you see this trick --- to add x23+x2x23+x2 \displaystyle \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} to the desired integral?

I know the question asks us to differentiate the functions but I still don't see this trick.

Then use ddx3+x2=... \frac{d}{dx}\sqrt{3 + x^2} = ... to get

3+x2dx+x23+x2x23+x2 \int \sqrt{3 + x^2} dx + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}}

Unparseable latex formula:

\displaystyle = x\sqrt{3 + x^2}- \left \int \frac{x^2}{\sqrt{3 + x^2}} \right


He seems to have used a less simplified form of the derivative of x3+x2x\sqrt{3+x^2}. Namely, by the product rule, ddx[x3+x2]=3+x2+x23+x2\dfrac{d}{dx}[x\sqrt{3+x^2}] = \sqrt{3+x^2} + \dfrac{x^2}{\sqrt{3+x^2}}. As soon as he saw that, it's obvious that he'll be able to evaluate an integral similar to that so he went for the addition-subtraction trick.

Second Question --- I know how to integrate x23+x2 -\int \frac{x^2}{\sqrt{3 + x^2}} by using the substitution x=atan(u) x = atan(u) . But the STEP solution says that no calculus work is needed for this part --- the integration of 3+x2dx \int \sqrt{3 + x^2} dx . I don't see the derivatives in the first part help me to get x23+x2 \displaystyle -\int \frac{x^2}{\sqrt{3 + x^2}}?

Thanks a lot---

He then noted a different form of the above derivative: ddx[x3+x2]=3+2x23+x2\dfrac{d}{dx}[x\sqrt{3+x^2}] = \dfrac{3+2x^2}{\sqrt{3+x^2}} (1)

and also:
ddx[ln(x+3+x2)]=13+x2\dfrac{d}{dx}[\ln(x+\sqrt{3+x^2})] = \dfrac{1}{\sqrt{3+x^2}} (2)

Furthermore:
x23+x2=12(3+2x23+x233+x2)\dfrac{x^2}{\sqrt{3+x^2}} = \dfrac{1}{2}\left(\dfrac{3+2x^2}{\sqrt{3+x^2}} - \dfrac{3}{\sqrt{3+x^2}}\right)
=12(ddx[x3+x2]3ddx[ln(x+3+x2)])= \dfrac{1}{2}\left(\dfrac{d}{dx}[x\sqrt{3+x^2}] -3\dfrac{d}{dx}[\ln(x+\sqrt{3+x^2})]\right)

by (1) and (2), which is easy to integrate. He made a mistake in the last line of working - There's a small sign error.

I have to say, I think this is the long way of doing it. When I did the question, I noted that:

ddx(x3+x2)=23+x233+x2\dfrac{d}{dx}(x\sqrt{3+x^2}) = 2\sqrt{3+x^2} - \dfrac{3}{\sqrt{3+x^2}}
    23+x2=ddx(x3+x2)+33+x2\iff 2\sqrt{3+x^2} = \dfrac{d}{dx}(x\sqrt{3+x^2}) + \dfrac{3}{\sqrt{3+x^2}}

Using (2), the RHS is easy to integrate and you're done.
(edited 11 years ago)
Reply 2
Great! Thanks ---
Reply 3
Original post by Farhan.Hanif93
I have to say, I think this is the long way of doing it. When I did the question, I noted that:

ddx(x3+x2)=23+x233+x2\dfrac{d}{dx}(x\sqrt{3+x^2}) = 2\sqrt{3+x^2} - \dfrac{3}{\sqrt{3+x^2}}
    23+x2=ddx(x3+x2)+33+x2\iff 2\sqrt{3+x^2} = \dfrac{d}{dx}(x\sqrt{3+x^2}) + \dfrac{3}{\sqrt{3+x^2}}

Using (2), the RHS is easy to integrate and you're done.


actually, one last question ---

How did you see the above? It wasn't and is still not obvious to me. The only way I see it is ---

3+2x23+x2=2(x2+3)3x2+3=2(x2+3)x2+33x2+3 \dfrac{3 + 2x^2}{\sqrt{3 + x^2}} = \dfrac{2(x^2 + 3) - 3}{\sqrt{x^2 + 3}} = \dfrac{2(x^2 + 3)}{\sqrt{x^2 + 3}} - \dfrac{3}{\sqrt{x^2 + 3}}

and it looks like only to be a very clever algebra trick ---

Thanks again ---
(edited 11 years ago)
Original post by adrienne_om
actually, one last question ---

How did you see the above? It wasn't and is still not obvious to me. The only way I see it is ---

2+3x23+x2=3(x2+2)3x2+3=3(x2+2)x2+33x2+3 \dfrac{2 + 3x^2}{\sqrt{3 + x^2}} = \dfrac{3(x^2 + 2) - 3}{\sqrt{x^2 + 3}} = \dfrac{3(x^2 + 2)}{\sqrt{x^2 + 3}} - \dfrac{3}{\sqrt{x^2 + 3}}

and it looks like only to be a very clever algebra trick ---

Thanks again ---

I think you've made a small typo at the start; it should be 3+2x2x2+3\dfrac{3+2x^2}{\sqrt{x^2+3}}. My reasoning behind it was that I wanted to rewrite this derivative in terms of 3+x2\sqrt{3+x^2} AND a constant multiple of the same or other derivative. It was clear to me that 2+3x23+x2\dfrac{2 + 3x^2}{\sqrt{3 + x^2}} can be rewritten in that form so I did the working and it just fell out.
(edited 11 years ago)
Original post by adrienne_om


How I did it was as follows:

ddx(ln(x+x2+3))=13(x2+3x2x2+3)\dfrac{d}{dx} \left( \ln(x+\sqrt{x^2+3}) \right) = \frac{1}{3} \left( \sqrt{x^2+3} - \dfrac{x^2}{\sqrt{x^2+3}} \right)

and ddx(x3+x2)=x2+3+x2x2+3\dfrac{d}{dx} \left( x \sqrt{3 + x^2} \right) = \sqrt{x^2 + 3} + \dfrac{x^2}{\sqrt{x^2+3}}


Now we can easily see that x2+3=12(x3+x2+3ln(x+x2+3))\displaystyle \int \sqrt{x^2 + 3} = \dfrac{1}{2} \left( x \sqrt{3 + x^2} + 3 \ln(x+\sqrt{x^2+3}) \right)
Reply 6
Thank you very much ---

I've also fixed my typo.

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