HSC 2016 MX2 Integration Marathon (archive) (4 Viewers)

leehuan · Nov 5, 2015

I do not feel at home without this...

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Here post integrals for helping 2016 students with integration!

Do not post a question until the last one is answered

(Reference to Sy123 for the copy and paste)
------------------------------------------------------------
FIRST QUESTION:

$\int { { x }^{ 3 }\sqrt { { x }^{ 2 }+1 } dx }$

InteGrand · Nov 5, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I do not feel at home without this...

------------------------------------------------------------
Here post integrals for helping 2016 students with integration!

Do not post a question until the last one is answered

(Reference to Sy123 for the copy and paste)
------------------------------------------------------------
FIRST QUESTION:

$\int { { x }^{ 3 }\sqrt { { x }^{ 2 }+1 } dx }$

The 2015 one still has some unanswered ones.

Ekman · Nov 5, 2015

Re: MX2 2016 Integration Marathon

InteGrand said:
The 2015 one still has some unanswered ones.

Then lets post them here, as the other one will most likely be archived

InteGrand · Nov 5, 2015

Re: MX2 2016 Integration Marathon

$Here are two currently unanswered ones from the user juantheron:$

juantheron said:
$(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx =$

$(3)\;\; If $F(n) = \int\frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

leehuan · Nov 5, 2015

Re: MX2 2016 Integration Marathon

I really couldn't be bothered doing the general case but I feel as though if I didn't treat c1=c2=0 things would've gone haywire. C3 doesn't seem as problematic though

Edit: So if we sub in the full integral and keep C3 there, our final result (using index laws) is going to be 66 * e^(C3), so we can just say the final answer is 66A for some positive arbitrary constant A.

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Paradoxica · Nov 6, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I really couldn't be bothered doing the general case but I feel as though if I didn't treat c1=c2=0 things would've gone haywire. C3 doesn't seem as problematic though

Edit: So if we sub in the full integral and keep C3 there, our final result (using index laws) is going to be 66 * e^(C3), so we can just say the final answer is 66A for some positive arbitrary constant A.

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Wolfram Alpha cannot evaluate the general case. I assume if one exists, it requires higher order functions found in real and complex analysis. So the question should be restated as...

$(3)\;\; If $F(n) = \int_0^x \frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

Anyway, my solution to 1.

juantheron said:
$(1)\;\; Evaluation of $\int_{0}^{\pi}\frac{x}{\sqrt{1+\sin^3 x}}((3\pi \cos x+4\sin x)\sin^2 x+4)dx =$

$(3)\;\; If $F(n) = \int\frac{x^n}{e^x}dx\;,$ Then value of Definite Integral $\int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$

$I = \int_0^{\pi} \frac{x}{\sqrt{1+\sin^3{x}}}\left ( 3\pi \cos{x} \sin^2{x} + 4\sin^3{x} + 4 \right )$

$$Consider $I_1 = \int_0^{\pi} \frac{x}{\sqrt{1+\sin^3{x}}} \left ( 3\pi \cos{x} \sin^2{x} \right ) dx$

$I_1 = \int_0^{\pi} \frac{\pi x d\left (1+\sin^3{x}\right )}{\sqrt{1+\sin^3{x}}} = \int_0^{\pi} 2\pi x d \left ( \sqrt{1+\sin^3{x}} \right )$

$$Using IBP, we get $ I_1 = \left [ 2\pi x\sqrt{1+\sin^3{x}}\right ] _0^{\pi} - 4 \int_0^{\pi} \frac{\pi}{2} \sqrt{1+\sin^3{x}} dx$

$I_1 = 2 \pi^2 - 4 \int_0^{\pi} \frac{\pi}{2} \sqrt{1+\sin^3{x}} dx$

$Substituting this back into $ I $, we have$

$I = 4 \int_0^{\pi} x \frac{1+\sin^3{x}}{\sqrt{1+\sin^3{x}}} dx - 4 \int_0^{\pi} \frac{\pi}{2} \sqrt{1+\sin^3{x}} dx + 2\pi^2$

$I = 4\int_0^{\pi} \left ( x - \frac{\pi}{2} \right ) \sqrt{1+ \sin^3{x}}dx + 2\pi^2 = 4\int_0^{\pi} \left (\frac{\pi}{2}-x \right ) \sqrt{1+ \sin^3{x}}dx + 2\pi^2$ (Symmetric border flipping)$$

$\therefore I = 2\pi^2$

leehuan · Nov 6, 2015

Re: MX2 2016 Integration Marathon

I'll get around to treating F(n) as that definite integral later. Anyway, hopefully some (non-accelerated...

) 2016ers will show up.

leehuan said:
$\int { { x }^{ 3 }\sqrt { { x }^{ 2 }+1 } dx }$

Paradoxica · Nov 6, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I'll get around to treating F(n) as that definite integral later. Anyway, hopefully some (non-accelerated... ) 2016ers will show up.

Oh, the definite form is identical to assuming the constants of integration are zero. It was just me editing the problem to fit your solution.

Technically I'm not accelerated so....

$I = \int x^3 \sqrt{x^2 + 1}dx = \int x \left ( x^2 + 1 \right )\sqrt{x^2 + 1} dx - \int x\sqrt{x^2 + 1} dx$

$= \frac{1}{2}\int d\left (x^2 +1 \right ) \left ( x^2 +1 \right )^{\frac{3}{2}} dx - \frac{1}{2}\int d\left (x^2 +1 \right ) \left ( x^2 +1 \right )^{\frac{1}{2}} dx$

The rest is done by substitution, yielding

$\left ( x^2 + 1 \right )\sqrt{x^2 +1}\left ( \frac{3x^2 -2}{15} \right ) + C$

calamebe · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
Technically I'm not accelerated so....

Are you doing your hsc in 2016, but have been tutored a year ahead?

Paradoxica · Nov 6, 2015

Re: MX2 2016 Integration Marathon

calamebe said:
Are you doing your hsc in 2016, but have been tutored a year ahead?

The tutoring helps with content digestion, but not with problem solving skills. I almost exclusively never use any of Truong's techniques. And if I do, that's because I see no other way, or am unaware of it.

DatAtarLyfe · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Apart from the accelerated kids, pretty much none of us are up to integration, so hoping that some of the 2016ers will answer is a little ambitious (unless of course some it's just harder 3u)

Sien · Nov 6, 2015

Re: MX2 2016 Integration Marathon

DatAtarLyfe said:
Apart from the accelerated kids, pretty much none of us are up to integration, so hoping that some of the 2016ers will answer is a little ambitious (unless of course some it's just harder 3u)

Integration is just backwards differentiation
Edit: crap didn't realise this was mx2 territory

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leehuan · Nov 6, 2015

Re: MX2 2016 Integration Marathon

I'm trying to keep the difficulty low for now. I'll increase it in February.

$\int { \sqrt { x } { e }^{ \sqrt { x } }dx }$

Most of these integrals should be doable with textbook browsing.

KingOfActing · Nov 6, 2015

Re: MX2 2016 Integration Marathon

leehuan said:
I'm trying to keep the difficulty low for now. I'll increase it in February.

$\int { \sqrt { x } { e }^{ \sqrt { x } }dx }$

Most of these integrals should be doable with textbook browsing.

$\begin{align*}\text{Let } u^2 = x \Longleftrightarrow dx = 2udu\\\int{\sqrt{x}e^{\sqrt{x}}dx}&=2\int{u^2e^udu}\\ &= 2(u^2e^u-2\int{ue^udu}) \text{ (By integration by parts)}\\ &= 2(u^2e^u - 2(ue^u-e^u))+C \text{ (By integration by parts)}\\ &= 2e^u(u^2-2u+2) + C\\ &= 2e^{\sqrt{x}}(x-2\sqrt{x}+2) + C\end{align*}$

NEXT QUESTION:
$\int{xe^{x}\sin{x}dx}$

Sy123 · Nov 6, 2015

Re: MX2 2016 Integration Marathon

$\\ \int_0^{2\cos \theta} \sqrt{x^2 - 2x \cos \theta + 1} \ $d$x$

$\\ $Take$ \ \theta \ $to be constant$$

outlier1 · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
The tutoring helps with content digestion, but not with problem solving skills. I almost exclusively never use any of Truong's techniques. And if I do, that's because I see no other way, or am unaware of it.

What's your opinion and rate on Truong's tutoring? Everyone says they're the best

Paradoxica · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ \int_0^{2\cos \theta} \sqrt{x^2 - 2x \cos \theta + 1} \ $d$x$

$\\ $Take$ \ \theta \ $to be constant$$

why must you taunt me?

$$Let $\tan{u} = x - \cos{\theta}$

$The integral is now similar to that of secant cubed, which can be done using IBP or reduction.$

Sy123 · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Paradoxica said:
why must you taunt me?

$$Let $\tan{u} = x - \cos{\theta}$

$The integral is now similar to that of secant cubed, which can be done using IBP or reduction.$

Not intentional haha

That substitution requires a constant in front of it, otherwise we can't simplify it to eliminate the square-root

KingOfActing · Nov 6, 2015

Re: MX2 2016 Integration Marathon

Sy123 said:
$\\ \int_0^{2\cos \theta} \sqrt{x^2 - 2x \cos \theta + 1} \ $d$x$

$\\ $Take$ \ \theta \ $to be constant$$

$Let $x = \sin{\theta}\tan{u}+\cos{\theta}\\$Our integral becomes $ \sin^2{\theta}\int_{-\tan^{-1}{(\cot{\theta})}}^{\tan^{-1}{(\cot{\theta})}}{\sec^3{u}du}\\$Since $ \sec^3{u} $ is an even function, our integral can be rewritten as $ 2\sin^2{\theta}\int_0^{\tan^{-1}{\cot{\theta}}}{\sec^3{u} d u}\\$This can then be evaluated using integration by parts, and simplified by using trigonometric identities.$

Paradoxica · Nov 6, 2015

Re: MX2 2016 Integration Marathon

outlier1 said:
What's your opinion and rate on Truong's tutoring? Everyone says they're the best

It's good if you're not willing to put the work to understand the syllabus content for mathematics, since he makes a lot of things one-size fits all techniques. This doesn't really help me though, as I consider everything on a case-by-case basis.

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