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

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{EASY QUESTION}$^\dagger$$$

$\noindent Find $\int \frac{1}{\sqrt{1+\mathrm{e}^{x}}}\text{ d}x$.$

$\noindent $^\dagger$ \small{Relative to most questions in this Marathon.}$

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
But I thought he did that already and then used the fundamental theorem of calculus to revert the derivative?

Or was the wrong part of the working what was being integrated. I'm not fully processing maths today to figure it out myself

His second last part is wrong then.

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent \textbf{EASY QUESTION}$^\dagger$$$

$\noindent Find $\int \frac{1}{\sqrt{1+\mathrm{e}^{x}}}\text{ d}x$.$

$\noindent $^\dagger$ \small{Relative to most questions in this Marathon.}$

$$ Let $ x=\ln{u} $ and then $ let v^2=1+u$

omegadot · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
His second last part is wrong then.

$\noindent No, I believe everything is correct.$

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent No, I believe everything is correct.$

Where did the -1 come from ?

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ Let $ x=\ln{u} $ and then $ let v^2=1+u$

$Correct!$

leehuan · Jan 12, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent No, I believe everything is correct.$

How does -1 become -x when the fundamental theorem of calculus is used?

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
How does -1 become -x when the fundamental theorem of calculus is used?

Wasn't the –1 just a typo? (By the way, this wasn't the fundamental theorem of calculus. The symbol ∫ on its own (without any limits of integration) just means an antiderivative, so ∫d/dx (f(x)) dx = f(x) (+C) simply because ∫ and d/dx are essentially inverse operators.)

omegadot · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Where did the -1 come from ?

$\noindent I just added the term $-1 + 1$ as Paradoxica suggested when the question was posed.$

omegadot · Jan 12, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
How does -1 become -x when the fundamental theorem of calculus is used?

InteGrand said:
Wasn't the –1 just a typo? (By the way, this wasn't the fundamental theorem of calculus. The symbol ∫ on its own (without any limits of integration) just means an antiderivative, so ∫d/dx (f(x)) dx = f(x) (+C) simply because ∫ and d/dx are essentially inverse operators.)

$$\noindent Oh, now I see what you mean! Yes, you are right. Have corrected this in my solution. It was a damn typo.$

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{NEW EASY QUESTION}$

$\noindent Find $\int \frac{\sqrt{1+x}}{x}\text{ d}x$.$

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Oh, now I see what you mean! Yes, you are right. Have corrected this in my solution. It was a damn typo.$

My heart's broken...

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
My heart's broken...

Why?

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Why?

Omega told me off for no reason...

New Question

$\int{\sqrt{\sin ^{-1}\sqrt{x}}}dx$

InteGrand · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Omega told me off for no reason...

New Question

$\int{\sqrt{\sin ^{-1}\sqrt{x}}}dx$

$\noindent I don't think omegadot was telling you off $\ddot{\smile}$.$

leehuan · Jan 12, 2016

Re: MX2 2016 Integration Marathon

${ u }^{ 2 }=1+x\Rightarrow 2u\, du=dx\\ \int { \frac { \sqrt { 1+x } }{ x } dx } \\ =\int { \frac { u }{ { u }^{ 2 }-1 } \left( 2u\, du \right) } \\ =\int { \left( 2+\frac { 2 }{ { u }^{ 2 }-1 } \right) du } \\ =\int { \left( 2+\frac { 1 }{ { u }-1 } -\frac { 1 }{ u+1 } \right) du } \\ =2\sqrt { 1+x } +\ln { \left| \frac { \sqrt { x+1 } -1 }{ \sqrt { x+1 } +1 } \right| } +C$

leehuan · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Omega told me off for no reason...

New Question

$\int{\sqrt{\sin ^{-1}\sqrt{x}}}dx$

http://www.wolframalpha.com/input/?i=integrate+sqrt(arcsin(sqrtx))

What.

seanieg89 · Jan 13, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
On the easy side, but:

$\lim_{R\rightarrow\infty}\int_{-R}^R \frac{\sin(ax)\sin(bx)}{x^2}\, dx \qquad (a,b\in\mathbb{R}).$

Bump.

Drsoccerball · Jan 13, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
http://www.wolframalpha.com/input/?i=integrate+sqrt(arcsin(sqrtx))

What.

It requires 3 substitutions.

KingOfActing · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
It requires 3 substitutions.

Are you sure it has a nice closed form? My first substitution was $x = \sin^2{(u^2)}$ which made the integral to to $\frac{-u}{2}\cos(2u^2)+\frac{1}{2}\int\cos(2u^2)\,du$ by integration by parts, and I'm pretty sure that second integral doesn't have a closed form solution.

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