# HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

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#### leehuan

##### Well-Known Member
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)
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FIRST QUESTION:

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

#### InteGrand

##### Well-Known Member
Re: MX2 2016 Integration Marathon

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:

$\bg_white \int { { x }^{ 3 }\sqrt { { x }^{ 2 }+1 } dx }$
The 2015 one still has some unanswered ones.

#### Ekman

##### Well-Known Member
Re: MX2 2016 Integration Marathon

The 2015 one still has some unanswered ones.
Then lets post them here, as the other one will most likely be archived

#### InteGrand

##### Well-Known Member
Re: MX2 2016 Integration Marathon

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

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

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

#### leehuan

##### Well-Known Member
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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##### -insert title here-
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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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...

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

Anyway, my solution to 1.

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

$\bg_white (3)\;\; If F(n) = \int\frac{x^n}{e^x}dx\;, Then value of Definite Integral \int_{2}^{5}\left[e^{\int\frac{F(1)}{F(2)}dx}\right]^2dx=$
$\bg_white I = \int_0^{\pi} \frac{x}{\sqrt{1+\sin^3{x}}}\left ( 3\pi \cos{x} \sin^2{x} + 4\sin^3{x} + 4 \right )$

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

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

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

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

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

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

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

$\bg_white \therefore I = 2\pi^2$

#### leehuan

##### Well-Known Member
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.

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

##### -insert title here-
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.
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....

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

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

The rest is done by substitution, yielding

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

#### calamebe

##### Active Member
Re: MX2 2016 Integration Marathon

Technically I'm not accelerated so....
Are you doing your hsc in 2016, but have been tutored a year ahead?

##### -insert title here-
Re: MX2 2016 Integration Marathon

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

##### Booty Connoisseur
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

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

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#### leehuan

##### Well-Known Member
Re: MX2 2016 Integration Marathon

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

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

Most of these integrals should be doable with textbook browsing.

#### KingOfActing

##### lukewarm mess
Re: MX2 2016 Integration Marathon

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

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

Most of these integrals should be doable with textbook browsing.
\bg_white \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:
$\bg_white \int{xe^{x}\sin{x}dx}$

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#### Sy123

##### This too shall pass
Re: MX2 2016 Integration Marathon

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

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

#### outlier1

##### New Member
Re: MX2 2016 Integration Marathon

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

##### -insert title here-
Re: MX2 2016 Integration Marathon

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

$\bg_white \\ Take \ \theta \ to be constant$
why must you taunt me?

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

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

#### Sy123

##### This too shall pass
Re: MX2 2016 Integration Marathon

why must you taunt me?

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

$\bg_white The integral is now similar to that of secant cubed, which can be done using IBP or reduction.$
Not intentional haha

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

#### KingOfActing

##### lukewarm mess
Re: MX2 2016 Integration Marathon

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

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