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

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
How did you p.f. it?

LOL

$I =\frac{-du}{(u+1)\sqrt{u^2+1}} = \frac{1}{2}\int ({\frac{u-1}{\sqrt{u^2+1}} - \frac{\sqrt{u^2+1}}{u+1}})du$
Yeah don't ask...

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Oh wow

Trig sub went right over your head

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Oh wow

Trig sub went right over your head

I've never done it before but it came in my mind for this question.

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I've never done it before but it came in my mind for this question.

Wait, you haven't??

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Wait, you haven't??

I havn't done p.f like that before.

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I havn't done p.f like that before.

Oh lol. Cause you quoted the post making it seem like you ignored trig sub.

But, um, wow m8

InteGrand · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Really? Cause whilst tedious this can be rearranged

Sent from my iPhone using Tapatalk

$\noindent In the radical in the second line, it should actually be $x^2 + 2+\frac{1}{x^2}\textbf{-1}$, not $+1$.$

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Oh lol. Cause you quoted the post making it seem like you ignored trig sub.

But, um, wow m8

I think my method would be faster if I knew inverse hyperbolic functions.

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent In the radical in the second line, it should actually be $x^2 + 2+\frac{1}{x^2}\textbf{-1}$, not $+1$.$

Strangely enough I had - in my head but wrote +. I even wondered "hm where did this + come from"

In that case, after the secant substitution...

$-\int { \frac { \sec { \theta } \tan { \theta } }{ \left( \sec { \theta } +1 \right) \sqrt { \sec ^{ 2 }{ \theta } -1 } } d\theta } \\ =-\int { \frac { \sec { \theta } }{ \left( \sec { \theta } +1 \right) } d\theta }$

This isn't even hard. A very nice result comes direct out of a tangent half-angle substitution

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Strangely enough I had - in my head but wrote +. I even wondered "hm where did this + come from"

In that case, after the secant substitution...

$-\int { \frac { \sec { \theta } \tan { \theta } }{ \left( \sec { \theta } +1 \right) \sqrt { \sec ^{ 2 }{ \theta } -1 } } d\theta } \\ =-\int { \frac { \sec { \theta } }{ \left( \sec { \theta } +1 \right) } d\theta }$

This isn't even hard. A very nice result comes direct out of a tangent half-angle substitution

$$\noindent Actually, by shifting the cosine from the denominator to the numerator, you get an integral that can be done by inspection.$ \\ I = \int \frac{\sec{x}}{1+\sec{x}} $d$x = \int \frac{1}{1+\cos{x}}$d$x = \int \frac{1-\cos{x}}{\sin^2{x}}$d$x = \int \csc^2{x} - \csc{x}\cot{x} $d$x \\ I = \csc{x}-\cot{x} + C$

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Im left with :

$\frac{1}{2}(\sqrt{(x+\frac{1}{x})^2+1} + \ln{(x+\frac{1}{x} +\sqrt{(x+\frac{1}{x})^2+1}})-\int{\frac{\sqrt{u^2+1}du}{u+1}})$

$$ Where -u $ = (x+\frac{1}{x})$

$Which is a non-elementary integral.$

That is not a non-elementary integral. I did the final form before through tedious manipulations.

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Actually, by shifting the cosine from the denominator to the numerator, you get an integral that can be done by inspection.$ \\ I = \int \frac{\sec{x}}{1+\sec{x}} $d$x = \int \frac{1}{1+\cos{x}}$d$x = \int \frac{1-\cos{x}}{\sin^2{x}}$d$x = \int \csc^2{x} - \csc{x}\cot{x} $d$x \\ I = \csc{x}-\cot{x} + C$

I considered it briefly, because I knew 1/(1±sinx) and 1/(1±cosx) were simple integrals that could be manipulated without the substitution using a Pythagorean identity. But being lazy to continue I checked on W.A. and you can just make it equal to -tan(x/2) anyhow.

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I think my method would be faster if I knew inverse hyperbolic functions.

Luckily, I already do

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Anyway, the answer is $\sqrt \frac{x^2+x+1}{x^2-x+1}+C$

$\noindent \mathbf{NEXT QUESTION}$

$\int \frac{1}{\sqrt{1-x} \sqrt{1-x+\sqrt{1-x}}} \, $d$x$

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Luckily, I already do

You got to teach me

(no joke pls teach me)

KingOfActing · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Anyway, the answer is $\sqrt \frac{x^2+x+1}{x^2-x+1}+C$

$\noindent \mathbf{NEXT QUESTION}$

$\int \frac{1}{\sqrt{1-x} \sqrt{1-x+\sqrt{1-x}}} \, $d$x$

$\begin{align*} &\int \frac{1}{\sqrt{1-x} \sqrt{1-x+\sqrt{1-x}}} \, dx \\&\text{Let } u = \sqrt{1-x}\\&\int \frac{-2}{\sqrt{u^2+u}}\,du \\&\text{Let } u + \frac{1}{2} = \frac{1}{2}\sec{\theta}\\&\int\frac{-\sec{\theta}\tan{\theta}}{\frac{1}{2}\tan{\theta}} \,d\theta = -2 \int \sec{\theta}\,d\theta = -2\ln\left|\sec\theta+\tan\theta\right| + C\\&=-2\ln|2u+1+\sqrt{4u^2+4u}|+ C\\&=-2\ln\left |2\sqrt{1-x}+1+2\sqrt{1-x +\sqrt{1-x}}\right | + C\end{align*}$

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Anyway, the answer is $\sqrt \frac{x^2+x+1}{x^2-x+1}+C$

$\noindent \mathbf{NEXT QUESTION}$

$\int \frac{1}{\sqrt{1-x} \sqrt{1-x+\sqrt{1-x}}} \, $d$x$

2 substitution leave the integral in the form of $\int{\sec{x}}dx$

In a rush

lel jokes beaten

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{1}{\left(\sqrt{x^2+ax}+x\right)^2} \, $d$x$

$\for a \in \mathbb{R}$

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{1}{\left(\sqrt{x^2+ax}+x\right)^2} \, $d$x$

$\for a \in \mathbb{R}$

After a couple of manipulations we get the integral
$I= \frac{1}{a}\ln{|x|} + \frac{2x}{a^2} -\frac{2}{a} \int{\frac{\sqrt{x+a}}{a\sqrt{x}}}dx$

$The remaining integral can be done with a substitution $ x= a \tan ^2 \theta $ Which give us the integral of $ \sec ^3 \theta $ which has been solved on these forums before.(AKA ceebs)$

InteGrand · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
After a couple of manipulations we get the integral
$I= \frac{1}{a}\ln{|x|} + \frac{2x}{a^2} -\frac{2}{a} \int{\frac{\sqrt{x+a}}{a\sqrt{x}}}dx$

$The remaining integral can be done with a substitution $ x= a \tan ^2 \theta $ Which give us the integral of $ \sec ^3 \theta $ which has been solved on these forums before.(AKA ceebs)$

$\noindent Note that this assumes $a\neq 0$. For the case of $a=0$, we have $I_0:=\int \frac{1}{\left(\sqrt{x^2}+x\right)^2}\text{ d}x$. Note that $x\neq 0$ as then the integrand has 0 denominator. In fact, $x\nless 0$ also, since if $x<0$, the denominator of the integrand is $-x+x=0$ again. So $x>0$ for this integral, so $\sqrt{x^2}=x$ and the integral is$

$\begin{align*}I_0 &= \int \frac{1}{\left(2x\right)^2}\text{ d}x \\ &= -\frac{1}{4x}+C, \, x>0.\end{align*}$

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