HSC 2016 MX2 Integration Marathon (archive) (5 Viewers)

juantheron · Jan 22, 2016

Re: MX2 2016 Integration Marathon

$let $I = \int\frac{1}{\left(x+\sqrt{x^2-1}\right)^{2}}dx$\;, Now put $(x+\sqrt{x^2-1}) = t\;,$

$Then $(x+\sqrt{x^2-1}) dx = \sqrt{x^2-1}dt$

$Now Using $(x+\sqrt{x^2-1})\cdot (x-\sqrt{x^2-1})=1 \;,$ Then $(x-\sqrt{x^2-1}) =\frac{1}{t}$

$So we get $\sqrt{x^2-1} = \frac{1}{2}\left(t-\frac{1}{t}\right) = \frac{t^2-1}{2t}$

$So $I = \frac{1}{2}\int \frac{t^2-1}{t^3}dt = \frac{1}{2}\ln |t|+\frac{1}{4}t^2+\mathcal{C}$

$So $I = \frac{1}{2}\ln \left|x+\sqrt{x^2-1}\right|+\frac{1}{4}\cdot \frac{1}{\left(x+\sqrt{x^2-1}\right)^2}+\mathcal{C}$

hedgehog_7 · Jan 22, 2016

Re: MX2 2016 Integration Marathon

integral of (cos x ) / (2 - cos x )

InteGrand · Jan 22, 2016

Re: MX2 2016 Integration Marathon

hedgehog_7 said:
integral of (cos x ) / (2 - cos x )

$\noindent We have$

$\begin{align*}\mathcal{I}:= \int \frac{\cos x}{2-\cos x }\text{ d}x &= -\int \frac{\cos x}{\cos x -2}\text{ d}x \\ &= -\int \frac{\cos x-2+2}{\cos x -2}\text{ d}x \\&= -\int \left(1+\frac{2}{\cos x -2} \right)\text{ d}x \\ &= -x - 2\int \frac{1}{\cos x -2}\text{ d}x. \end{align*}$

$\noindent This last integral can be done in a variety of ways, e.g. by using the substitution $t=\tan \frac{x}{2}$.$

Paradoxica · Jan 23, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$let $I = \int\frac{1}{\left(x+\sqrt{x^2-1}\right)^{2}}dx$\;, Now put $(x+\sqrt{x^2-1}) = t\;,$

$Then $(x+\sqrt{x^2-1}) dx = \sqrt{x^2-1}dt$

$Now Using $(x+\sqrt{x^2-1})\cdot (x-\sqrt{x^2-1})=1 \;,$ Then $(x-\sqrt{x^2-1}) =\frac{1}{t}$

$So we get $\sqrt{x^2-1} = \frac{1}{2}\left(t-\frac{1}{t}\right) = \frac{t^2-1}{2t}$

$So $I = \frac{1}{2}\int \frac{t^2-1}{t^3}dt = \frac{1}{2}\ln |t|+\frac{1}{4}t^2+\mathcal{C}$

$So $I = \frac{1}{2}\ln \left|x+\sqrt{x^2-1}\right|+\frac{1}{4}\cdot \frac{1}{\left(x+\sqrt{x^2-1}\right)^2}+\mathcal{C}$

Your solution to your problem isn't correct, and it's not the same as what Drsoccerball posted.

As for omegadot's question...

My euler substitution refuses to work.

Paradoxica · Jan 23, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Next Question}$

$$\noindent Find $\int \sqrt{x^2 + x^{-2} - 1} \, dx.$

$$\noindent$ I = \int \sqrt{x^4-x^2+1} \frac{\text{d}x}{x}= \frac{1}{2}\int \sqrt{v^2-v+1}\frac{\text{d}v}{v} \\ \sqrt{v^2-v+1} = v+t \Leftrightarrow v = \frac{1-t^2}{2t+1} \Rightarrow $d$v = -2 \frac{t^2+t+1}{(2t+1)^2}$d$t \\ v+t = \frac{t^2+t+1}{2t+1}; \; \frac{1}{v} = \frac{1-t^2}{2t+1} \\ I = \int \frac{(t^2+t+1)^2}{(t^2-1)(2t+1)^2}$d$t \\I= \int -\frac{1}{2 (t+1)}+\frac{1}{2 (2 t+1)}-\frac{3}{4 (2 t+1)^2}+\frac{1}{2 (t-1)}+\frac{1}{4}$d$t \\ 8I = 2 t+ 1 +\frac{3}{2 t+1}+4 \log (1-t)-4 \log (t+1)+2 \log (2 t+1) +C_0 \\ 8I = 4\frac{t^2 + t + 1}{2t+1} + 4\log{\left( \frac{1-t^2}{2t+1}\right )} - 4\log{\left (\frac{3(1+t)^2}{2t+1} \right )} -2\log{\left(\frac{3}{2t+1}\right)} +C_1 \\ 2I = (v+t) -\log{v} -\log{\left(\frac{2(t^2 +t +1) + t^2-1+2(2t+1)}{2t+1} \right)}-\frac{1}{2}\log{\left(\frac{2(1 - t^2) - (2t + 1) + 2(t^2 + t + 1)}{2t+1} \right)}$

$$\noindent$ 2I = v+t - \log{v} -\log{(2(v+t)+v+2)}-\frac{1}{2}\log(2v-1+2(v+t)) \\2I = \sqrt{\Delta}-2\log{x}-\log(2\sqrt{\Delta}-x^2+2)-\frac{1}{2}\log(2x^2-1+2\sqrt{\Delta}) \\ I = \frac{1}{2}\sqrt{x^4-x^2+1}-\log{x}-\frac{1}{2}\log(2\sqrt{x^4-x^2+1}-x^2+2)-\frac{1}{4}\log(2x^2-1+2\sqrt{x^4-x^2+1})$

Paradoxica · Jan 23, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{\sin ^{-1}\left(e^x\right)}{e^x} \, $d$x$

leehuan · Jan 23, 2016

Re: MX2 2016 Integration Marathon

$u={ e }^{ x }\Rightarrow du={ e }^{ x }dx\\ \int { \frac { \arcsin { { e }^{ x } } }{ { e }^{ x } } dx } \\ =\int { { u }^{ -2 }\arcsin { u } \, du } \\ =-\frac { \arcsin { u } }{ u } +\int { \frac { du }{ u\sqrt { \left( 1-{ u }^{ 2 } \right) } } } \\ u=\sin { \theta } \Rightarrow du=\cos { \theta } d\theta \\ =\int { \frac { \cos { \theta } du }{ \sin { \theta } \cos { \theta } } } -\frac { \arcsin { { e }^{ x } } }{ { e }^{ x } } \\ =-\ln { \left| \csc { \theta } +\cot { \theta } \right| } -\frac { \arcsin { { e }^{ x } } }{ { e }^{ x } } +C\\ =-\ln { \left| \frac { 1 }{ u } +\frac { \sqrt { 1-u^{ 2 } } }{ u } \right| } -\frac { \arcsin { { e }^{ x } } }{ { e }^{ x } } +C\\ =x -\ln { \left( 1+\sqrt { 1-{ e }^{ 2x } } \right) } -\frac { \arcsin { { e }^{ x } } }{ { e }^{ x } } +C$

Edit: Lmao just realised I wrote ln(e^x) and forgot to turn that back to x

Paradoxica · Jan 25, 2016

Re: MX2 2016 Integration Marathon

$\noindent $\frac{X\cos^2{\theta} + Y\sin^2{\theta} + Z\sin{\theta} \cos{\theta}}{a\cos{\theta}+b\sin{\theta}} \equiv p\cos{\theta}+q\sin{\theta}+\frac{r}{a\cos{\theta}+b\sin{\theta}} \\ $The above identity holds true for all fixed $\{a,b,X,Y,Z\} \in \mathbb{R}$, with $\{p,q,r\} \in \mathbb{R}$ all variable, subject to restrictions on the denominator.$ \\ $a) By combining the expression on the right hand side, determine $p,q,r$ in terms of $a,b,X,Y,Z$. (Hint: $\cos^2{\theta}+\sin^2{\theta}\equiv 1) \\$b) Given $a^2+b^2=c^2$, evaluate $\int \frac{X\cos^2{\theta} + Y\sin^2{\theta} + Z\sin{\theta} \cos{\theta}}{a\cos{\theta}+b\sin{\theta}} $d$ \theta \\ $The final condition just makes the auxiliary transformation nice, so there's nothing special about it.$

leehuan · Jan 25, 2016

Re: MX2 2016 Integration Marathon

I've just been lazy to do an integral lol I could kinda see where this was going.

hedgehog_7 · Jan 25, 2016

Re: MX2 2016 Integration Marathon

integrate 1/(x^2+2x-1)^(1/2)

Drsoccerball · Jan 25, 2016

Re: MX2 2016 Integration Marathon

hedgehog_7 said:
integrate 1/(x^2+2x-1)^(1/2)

You should learn Latex

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

Press "reply with Quote" to see what I typed.

InteGrand · Jan 25, 2016

Re: MX2 2016 Integration Marathon

hedgehog_7 said:
integrate 1/(x^2+2x-1)^(1/2)

$\noindent Complete the square, so the quantity inside the radical becomes $(x+1)^2 -2$. So the answer is $\ln \left(x+1 +\sqrt{(x+1)^2 -2}\right) + C = \ln \left(x+1 +\sqrt{x^2 + 2x-1\right) + C$, using the standard integrals sheet from past HSC papers.$

leehuan · Jan 25, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Complete the square, so the quantity inside the radical becomes $(x+1)^2 -2$. So the answer is $\ln \left(x+1 +\sqrt{(x+1)^2 -2}\right) + C = \ln \left(x+1 +\sqrt{x^2 + 2x-1\right) + C$, using the standard integrals sheet from past HSC papers.$

I have a bad feeling that (x+1)=tan(theta) is now required.

But yeah it does seem like he took a question out of a past paper so whatever

Drsoccerball · Jan 25, 2016

Re: MX2 2016 Integration Marathon

$$ (i) Prove that \sin{x} \cdot ( \sum_{r=1}^{2n-1}{\sin{(rx)}}) = \sin ^2 nx$

$$ (ii) Hence, find: $ \int{\frac{\sin ^2 4x}{\sin{x}}}dx$

marxman · Jan 25, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ (i) Prove that \sin{x} \cdot ( \sum_{r=1}^{2n-1}{\sin{(rx)}}) = \sin ^2 nx$

$$ (ii) Hence, find: $ \int{\frac{\sin ^2 4x}{\sin{x}}}dx$

Sorry I'm not familiar with latex...

Paradoxica · Jan 26, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{1-x^2}{\left(x^2-x+1\right) \sqrt{x^4+x^2+1}} \, dx$

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{1-x^2}{\left(x^2-x+1\right) \sqrt{x^4+x^2+1}} \, dx$

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})$

$Which is a non-elementary integral.$

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Really? Cause whilst tedious this can be rearranged

Sent from my iPhone using Tapatalk

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Hmmm I shouldnt have used partial fractions it made a non elementary integral.

leehuan · Jan 26, 2016

Re: MX2 2016 Integration Marathon

How did you p.f. it?

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