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

Drsoccerball · Jan 26, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\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$

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

So there's 2 cases one for a=0 and one for all other values ?

InteGrand · Jan 26, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
So there's 2 cases one for a=0 and one for all other values ?

Yeah.

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

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

InteGrand · Jan 27, 2016

Re: MX2 2016 Integration Marathon

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

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

$\noindent For $a\neq 0$, we can also do it as follows. Let $a\in \mathbb{R}$ and $a\neq 0$. Let $I:= \int \frac{1}{\left(\sqrt{x^2 + ax}+x\right)^2}\text{ d}x$. Make the substitution $z=\sqrt{x^2 + ax}+x$, so$

$\begin{align*}\sqrt{x^2 +ax}&= z-x \quad (\text{aside: this is known as an \textit{Euler Substitution}}) \\ \Rightarrow \color{red}{x^2}\color{black} + ax &= z^2 - 2zx \color{red}+ x^2 \color{black} \quad (\text{note that the red terms cancel}) \\ \Rightarrow (2z+a)x &= z^2 \\ \Rightarrow x &= \frac{z^2}{2z+a}.\end{align*}$

$\noindent So$

$\begin{align*}\frac{\mathrm{d}x}{\mathrm{d}z}&= \frac{2z(2z+a)-z^2 \cdot 2}{(2z+a)^2} \quad (\text{quotient rule}) \\ \Rightarrow \mathrm{d}x&= \frac{2z^2 + 2az}{(2z+a)^2}\text{ d}z.\end{align*}$

$\noindent Implementing these substitutions, we have$

$\begin{align*}I&= \int \frac{1}{z^2}\cdot \frac{2z^2 + 2az}{(2z+a)^2}\text{ d}z \\ &= \int \frac{2z+2a}{z(2z+a)^2}\text{ d}z \\ &= 2\int \frac{z+a}{2^2 \cdot z\left(z+\frac{a}{2}\right)^2}\text{ d}z \\ &= \frac{1}{2}\int \frac{z+a}{ z\left(z+\frac{a}{2}\right)^2}\text{ d}z \\ \Rightarrow 2I &= \int \frac{z+a}{ z\left(z+\frac{a}{2}\right)^2}\text{ d}z .\end{align*}$

$\noindent Now, let $\frac{z+a}{ z\left(z+\frac{a}{2}\right)^2} \equiv \frac{A}{z}+\frac{B}{z+\frac{a}{2}}+\frac{C}{\left(z+\frac{a}{2}\right)^2}$.$

$\noindent Then $z+a\equiv A\left(z+\frac{a}{2}\right)^2 + B z\left(z+\frac{a}{2}\right) +Cz$.$

$\noindent Substituting $z=-\frac{a}{2}$ yields $-\frac{a}{2}+a = -C\cdot \frac{a}{2} \Rightarrow \frac{a}{2}= -C\cdot \frac{a}{2} \Rightarrow \boxed{C=-1}$.$

$\noindent Substituting $z=0$ yields $0+a = A\cdot \left(\frac{a}{2}\right)^2 \Rightarrow a= A\times \frac{a^2}{4} \Rightarrow \boxed{A=\frac{4}{a}}$.$

$\noindent Equating the coefficients of $z^2$ yields $0=A+B \Rightarrow B=-A \Rightarrow \boxed{B=-\frac{4}{a}}$. So we have$

$\begin{align*}2I&= \int \left(\frac{\frac{4}{a}}{z}+\frac{-\frac{4}{a}}{z+\frac{a}{2}}+\frac{-1}{\left(z+\frac{a}{2}\right)^2} \right)\text{ d}z \\ &=\frac{4}{a}\ln \left|\frac{z}{z+\frac{a}{2}} \right| +\frac{1}{z+\frac{a}{2}}+c^\prime \\ \Rightarrow I&=\frac{2}{a}\ln \left|\frac{z}{z+\frac{a}{2}} \right| +\frac{1}{2z+a}+c \quad (\text{halving both sides, }c\text{ an arbitrary constant}) \\ &=\frac{2}{a}\ln \left|\frac{\sqrt{x^2 + ax}+x}{\sqrt{x^2 + ax}+x+\frac{a}{2}} \right| +\frac{1}{2\left(\sqrt{x^2 + ax}+x\right)+a}+c. \end{align*}$

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

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

omegadot · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\sqrt{1-x^2}}{x^2+1} \, $d$x$

$$\noindent Let $x = \sin \theta, dx = \cos \theta \, d\theta.$ So$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2+1} \, dx &= \int \frac{\cos^2 \theta}{\sin^2 \theta + 1} \, d\theta\\&=\int \frac{\cos^2 \theta}{\sin^2 \theta + (\sin^2 \theta + \cos^2 \theta)} \, d\theta\\&=\int \frac{\cos^2 \theta}{2 \sin^2 \theta + \cos^2 \theta} \, d\theta\\&= \int \frac{d\theta}{2 \tan^2 \theta + 1}\end{align*}$

$\noindent Now let $t = \tan \theta, dt = \sec^2 \theta \, d\theta \Rightarrow d\theta = \frac{dt}{1 + t^2}.$ Thus$

$\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2+1} \, dx &= \int \frac{dt}{(1 + t^2)(1 + 2t^2)}\\ &= \int \left (\frac{2}{1 + 2t^2} - \frac{1}{2t^2 + 1} \right ) \, dt\\ &= \sqrt{2} \tan^{-1} (\sqrt{2} t) - \tan^{-1} t + \cal{C}\\&=\sqrt{2} \tan^{-1} \left (\frac{\sqrt{2} x}{\sqrt{1 - x^2}} \right ) - \tan^{-1} \left (\frac{x}{\sqrt{1 - x^2}} \right ) + \cal {C}\end{align*}$

$\noindent \textbf{Next Question}$

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

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Let $x = \sin \theta, dx = \cos \theta \, d\theta.$ So$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2+1} \, dx &= \int \frac{\cos^2 \theta}{\sin^2 \theta + 1} \, d\theta\\&=\int \frac{\cos^2 \theta}{\sin^2 \theta + (\sin^2 \theta + \cos^2 \theta)} \, d\theta\\&=\int \frac{\cos^2 \theta}{2 \sin^2 \theta + \cos^2 \theta} \, d\theta\\&= \int \frac{d\theta}{2 \tan^2 \theta + 1}\end{align*}$

$\noindent Now let $t = \tan \theta, dt = \sec^2 \theta \, d\theta \Rightarrow d\theta = \frac{dt}{1 + t^2}.$ Thus$

$\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2+1} \, dx &= \int \frac{dt}{(1 + t^2)(1 + 2t^2)}\\ &= \int \left (\frac{2}{1 + 2t^2} - \frac{1}{2t^2 + 1} \right ) \, dt\\ &= \sqrt{2} \tan^{-1} (\sqrt{2} t) - \tan^{-1} t + \cal{C}\\&=\sqrt{2} \tan^{-1} \left (\frac{\sqrt{2} x}{\sqrt{1 - x^2}} \right ) - \tan^{-1} \left (\frac{x}{\sqrt{1 - x^2}} \right ) + \cal {C}\end{align*}$

Please simplify your answer. Exactly one of those expressions can be changed into a different inverse trigonometric function.

omegadot · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Please simplify your answer. Exactly one of those expressions can be changed into a different inverse trigonometric function.

$$\noindent Ah yes, the second one just equals $\sin^{-1} x$. So for completeness$\\\begin{align*}\int \frac{\sqrt{1-x^2}}{x^2 + 1} \, dx = \sqrt{2} \tan^{-1} \left ( \frac{\sqrt{2} x}{\sqrt{1 - x^2}} \right ) + \sin^{-1} x + \cal{C}\end{align*}$

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Find $\int \frac{x}{\sqrt{x^2 - 1} (\sqrt{x^2 - 1} + x)^2} \, dx.$

$$\noindent$ I = \int \frac{x\text{d}x}{\sqrt{x^2 -1} (\sqrt{x^2 -1} + x)^2} = \int \frac{\text{d}(\sqrt{x^2-1})}{(x+\sqrt{x^2-1})^2} = \frac{\sqrt{x^2-1}}{(x+\sqrt{x^2-1})^2} + 2\int \frac{\sqrt{x^2-1}(1+\frac{x}{\sqrt{x^2-1}})\text{d}x}{(x+\sqrt{x^2-1})^3}=\frac{\sqrt{x^2-1}}{(x+\sqrt{x^2-1})^2} + 2\int \frac{\text{d}x}{(x+\sqrt{x^2-1})^2}\\ I = \frac{\sqrt{x^2-1}}{(x+\sqrt{x^2-1})^2}+\frac{2x}{(x+\sqrt{x^2-1})^2}+4\int \frac{x(1+\frac{x}{\sqrt{x^2-1}})\text{d}x}{(x+\sqrt{x^2-1})^3} \\ I = \frac{2x+\sqrt{x^2-1}}{(x+\sqrt{x^2-1})^2}+4\int \frac{x\text{d}x}{\sqrt{x^2-1}(x+\sqrt{x^2-1})^2} \\ I = \frac{2x+\sqrt{x^2-1}}{(x+\sqrt{x^2-1})^2} +4I \\ I = -\frac{2x+\sqrt{x^2-1}}{3(x+\sqrt{x^2-1})^2}+C$

Drsoccerball · Jan 27, 2016

Re: MX2 2016 Integration Marathon

$\int_0^{\pi}{\frac{\sin{(2015x) dx}}{\sin{x}}}$

leehuan · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$\int_0^{\pi}{\frac{\sin{(2015x) dx}}{\sin{x}}}$

I swear I did this question last year...

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I swear I did this question last year...

Well, looking at this question, I'm thinking reduction for odd multiples of x, and then using the fact that the smallest positive odd number integrates to pi, all of the odd multiples of x integrate to pi.

Also, the even ones integrate to zero, but you knew that.

Drsoccerball · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Well, looking at this question, I'm thinking reduction for odd multiples of x, and then using the fact that the smallest positive odd number integrates to pi, all of the odd multiples of x integrate to pi.

Also, the even ones integrate to zero, but you knew that.

We could use the formula we proved on the previous page.

leehuan · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Well, looking at this question, I'm thinking reduction for odd multiples of x, and then using the fact that the smallest positive odd number integrates to pi, all of the odd multiples of x integrate to pi.

Also, the even ones integrate to zero, but you knew that.

I wrote the reduction integral to In=\int sin(nx)/sin(x) last year.

Lol I actually DID know the even ones went to 0.

InteGrand · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
We could use the formula we proved on the previous page.

Which formula did we prove on the previous page? Can't find any relevant ones there.

omegadot · Jan 27, 2016

Re: MX2 2016 Integration Marathon

$\noindent Need to keep the train moving right along with a \textbf{New Question}$

$$\noindent Evaluate $\int^1_0 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, dx.$

Drsoccerball · Jan 27, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Which formula did we prove on the previous page? Can't find any relevant ones there.

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$

.

Drsoccerball · Jan 27, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent Need to keep the train moving right along with a \textbf{New Question}$

$$\noindent Evaluate $\int^1_0 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, dx.$

We can make use of my partial fractions formula :

$I=\frac{1}{2}\int_0^1{\frac{1}{\sqrt{u+2}}- \frac{\sqrt{u+2}}{(u+2)+1}}dx$

$$ After $ u=x^2$

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
We can make use of my partial fractions formula :

$I=\frac{1}{2}\int_0^1{\frac{1}{\sqrt{u+2}}- \frac{\sqrt{u+2}}{(u+2)+1}}$d$x$

$$ After $ u=x^2$

lel noob

$$\noindent$ v = \sqrt{x^2 +2} \Leftrightarrow $d$v = \frac{x\text{d}x}{\sqrt{x^2+2}} \, , \, \, 0 \mapsto \sqrt{2} \, , \, \, 1 \mapsto \sqrt{3} \\ \int^1_0 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, $d$x = \int_{\sqrt{2}}^{\sqrt{3}} \frac{\text{d}v}{v^2+1} \\ \int_0^1 \frac{x}{(x^2 + 3) \sqrt{x^2 + 2}} \, $d$x = \frac{\pi}{3}-\tan^{-1}\sqrt{2}$

Paradoxica · Jan 27, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$\int_0^{\pi}{\frac{\sin{(2015x) dx}}{\sin{x}}}$

$$\noindent Consider the generalised integral $I_{2n+1} = \int_0^\pi \frac{\sin{(2k+1)x}}{\sin{x}}$d$ x \\ $Applying the angle sum expansion to $(2k-1)x + 2x \equiv (2k+1)x$ yields:$ \\ I = \int_0^\pi \frac{\sin{(2k-1)x} \cos{2x}+\cos{(2k-1)x} \sin{2x}}{\sin{x}} \\ I = \int_0^\pi \frac{(1-2\sin^2{x})\sin{(2k-1)x}}{\sin{x}} + \frac{2\sin{x}\cos{x}\cos{(2k-1)x} }{\sin{x}} \, $d$x \\$The second term in the integrand vanishes due to the parity of $\cos{nx} \, \forall n \in \mathbb{N}$ around $ x = \frac{\pi}{2} \\ $Expanding out the left term, we have $ I = \int_0^\pi \frac{\sin{(2k-1)x}}{\sin{x}} - 2\sin{x}\sin{(2k-1)x} \, $d$x \\ $The new second term can be re-written using the product to sum formulas in terms of the cosine function, and once again, it vanishes due to parity. This leaves us with a similar integral with integer multiple shifted down by two.$

$$\noindent So we conclude $I_{2n+1}=I_{2n-1}$, and iterating this arbitrarily many times, we conclude $I_{2n+1} = I_1 \, \forall n \in \mathbb{N}. \\ I_1 = \int_0^\pi \frac{\sin{x}}{\sin{x}} $d$x = \int_0^\pi $d$x = \pi \\ \therefore \int_0^\pi \frac{\sin{2015x}}{\sin{x}} $d$x = \pi$

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