HSC 2016 MX2 Integration Marathon (archive) (11 Viewers)

Paradoxica · Feb 29, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
Had a quick crack at the integral with the nested square roots in my break.

So the main difficulty in the problem is handling the integrand, it is not at all clear that this formally written object will be a well defined function, and on which domain it will be defined. (These sorts of things can also have different expressions in different parts of their domain).

If we denote this function by g(x), then from the positivity of square roots, we expect that g is a function defined on the interval of integration (at least) that satisfies:

i) (g(x)^2-x)^2=x - g(x)
ii) x < g(x)^2 < x + sqrt(x)

I believe that these conditions uniquely determine the function g on this interval, given by g(x)=(1+sqrt(4x-3))/2.

This function is straightforward to integrate and gives 127/6.

(Upon confirmation of this answer, I will post how I obtained that expression for g(x) later, but until then I won't ruin it for others. I rushed this calculation a bit so an inaccuracy would not surprise me.)

Correct.

seanieg89 · Feb 29, 2016

Re: MX2 2016 Integration Marathon

Cool, I will post my derivation of the expression for g(x) after my supervisor meeting tomorrow. Would like to take the time to explain it clearly.

The main idea though:

Check out the equation relating g(x) and x. It is quartic in g(x) but only quadratic in x.

This means we can write x fairly simply in terms of g(x). This ends up only being a quadratic polynomial in g(x) due to a pleasant cancellation.

We can then solve this quadratic for g(x) telling us what g(x) can be in terms of x. ("Can be" because solving the two quadratics leads to two independent plus-minus signs, and even if we were to narrow down these possibilities we are not guaranteed success, because we don't know that the solution g we are looking for exists a priori.)

Anyway, the two inequalities mentioned in my previous post actually force the choices of signs, and for these choices, the resulting function g(x)=(1+sqrt(4x-3))/2 satisfies the desired properties.

juantheron · Mar 1, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\int{\left(\sqrt{\tan{x}}+\sqrt{\cot{x}}\right) dx}$

$Let $I = \int\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx = \sqrt{2}\int\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

$Now put $\sin x-\cos x = t\Rightarrow 1-t^2=\sin 2x\;,$ Then $(\cos x+\sin x)dx = dt$

$So Integral $I = \sqrt{2}\int\frac{1}{\sqrt{1-t^2}}dt = \sqrt{2}\sin^{-1}(t)+\mathcal{C}$

$So $I = \int\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}dx = \sqrt{2}\sin^{-1}\left(\sin x-\cos x\right)+\mathcal{C}$

juantheron · Mar 1, 2016

Re: MX2 2016 Integration Marathon

$(1)\; $Evaluate $I = \int_{0}^{1}4x^3\left\{\frac{d^2}{dx^2}(1-x^2)^5\right\}dx$$

$(2)\;$If $I = \int_{0}^{4\pi}e^t\left(\sin^6 t+\cos^4 t\right)dt$ and $J = \int_{0}^{\pi}e^t\left(\sin^6 t+\cos^4 t\right)dt\;,$$

$Then $\frac{I}{J}\;,$Where $a\in \mathbb{N}$

Paradoxica · Mar 1, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$(1)\; $Evaluate $I = \int_{0}^{1}4x^3\left\{\frac{d^2}{dx^2}(1-x^2)^5\right\}dx$$

$(2)\;$If $I = \int_{0}^{4\pi}e^t\left(\sin^6 t+\cos^4 t\right)dt$ and $J = \int_{0}^{\pi}e^t\left(\sin^6 t+\cos^4 t\right)dt\;,$$

$Then $\frac{I}{J}\;,$Where $a\in \mathbb{N}$

$$\noindent The first integral can be done using two applications of Integration By Parts in order to avoid dealing with the second derivative of the expression given. The first two terms vanish over the integration bounds, and the final term evaluates to $2$. Thus, the integral is equal to $2$.$

$\noindent$I$ can be subdivided into four integrals that are all similar to $J$. By substitution or simple observation, each of the four integrals are proportional to $J$. The first subdivision is identical to $J$, the second is $e^\pi$ times as much as $J$, the third is $e^{2\pi}$ times as much as $J$, and the fourth is $e^{3\pi}$ times as much as $J$. Thus, the ratio of $I$ to $J$ is $e^{3\pi} + e^{2\pi} + e^\pi + 1 = \frac{e^{4\pi}-1}{e^\pi -1}\\$Remark: The above process can be continued inductively as to extend the property to arbitrarily large multiples of $\pi$.$

omegadot · Mar 1, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent b) Hence evaluate $\int \frac{2x-3}{x(x-1)(x-2)(x-3)+2}$

$\begin{align*}\int \frac{2x - 3}{x(x - 1)(x- 2)(x - 3) + 2} \, dx &= \int \frac{2x - 3}{x (x - 3) \cdot (x - 2)(x - 3) + 2} \, dx\\&= \int \frac{2x - 3}{(x^2 - 3x)(x^2 - 3x + 2) + 2} \, dx\end{align*}$

$$\noindent Let $u = x^2 - 3x, du = (2x - 3) \, dx$ so$

$\begin{align*}\int \frac{2x - 3}{x(x - 1)(x- 2)(x - 3) + 2} \, dx &= \int \frac{du}{u (u + 2) + 2}\\ &= \int \frac{du}{u^2 + 2u + 2}\\ &= \int \frac{du}{(u + 1)^2 + 1}\\ &= \tan^{-1} (u + 1) + \cal{C}\\&= \tan^{-1} (x^2 - 3x + 1) + \cal{C}\end{align*}$

omegadot · Mar 1, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Can someone give me a hint in this : $\int_0^1{\frac{1}{(x^2-x+1)(e^{2x-1}+1)}}dx$

Nevermind got it.

$$\noindent In case anyone would like to see how this one is done.$

$$\noindent Let $I = \int_0^1{\frac{1}{(x^2-x+1)(e^{2x-1}+1)}} \, dx$

$\noindent Using a so-called ``boarder flip'' we have:$

$\begin{align*}I &= \int^1_0 \frac{dx}{[(1 - x)^2 - (1 - x) + 1][e^{2(1 - x)} + 1]}\\ &= \int^1_0 \frac{dx}{(x^2 - x + 1)(e^{1 - 2x} + 1)}\\ \Rightarrow I + I &= \int^1_0 \left (\frac{1}{e^{2x - 1} + 1} + \frac{1}{e^{1 - 2x} + 1} \right ) \frac{dx}{x^2 - x + 1}\\ 2I &= \int^1_0 \frac{dx}{x^2 - x + 1}\\ \Rightarrow I &= \frac{1}{2} \int^1_0 \frac{dx}{(x - \frac{1}{2})^2 + \frac{3}{4}}\\&= \frac{1}{\sqrt{3}} \tan^{-1} \left (\frac{2x - 1}{\sqrt{3}} \right ) \Big{|}^1_0\\ &= \frac{\pi}{3 \sqrt{3}}\end{align*}$

omegadot · Mar 1, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^5_0 \frac{|x - 1|}{|x - 2| + |x - 4|} \, dx.$

Paradoxica · Mar 1, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Evaluate $\int^5_0 \frac{|x - 1|}{|x - 2| + |x - 4|} \, dx.$

such inelegance

The integral can be split along the abscissae 1,2,4. Taking cases will then allow the resultant rational functions to be simplified.

$$\noindent This evaluates to $\frac{5}{2} + 3\log{2} - \log{3}$

Surely there must be a more elaborate means of computing this area.

juantheron · Mar 1, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Evaluate $ \int_{0}^\infty \frac{\log{x}\,\text{d}x}{(x + a)^2 + b^2} $ by using the substitution $ x = \frac{a^2+b^2}{y}\\ $It is given that $\{a,b\}\in \mathbb{R}^+ \\ $Hence evaluate $\int_0^\infty \frac{\log{x}\, \text{d}x}{x^2 + 2ax + 2a^2}$

$$Put $x=\frac{a^2+b^2}{y}\;,$ Then $dx = -\frac{a^2+b^2}{y^2}$ and changing Limits,We get$

$I = \int_{\infty}^{0}\frac{\ln\left(\frac{a^2+b^2}{y}\right)}{\left(\frac{a^2+b^2}{y}+a\right)^2+b^2}\times -\frac{(a^2+b^2)}{y^2}dy$

$So $I = \int_{0}^{\infty}\frac{\ln(a^2+b^2)-\ln y}{(y+a)^2+b^2}dy$

$So $I =\int_{0}^{\infty}\frac{\ln(a^2+b^2)-\ln x}{(x+a)^2+b^2}dx$

$Above we have used $\int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$ and $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(t)dt$

$So $I = \ln(a^2+b^2)\int_{0}^{\infty}\frac{1}{(x+a)^2+b^2}dx-I$

$So $2I = \ln(a^2+b^2)\cdot \frac{1}{b}\cdot \left[\tan^{-1}\left(\frac{x+a}{b}\right)\right]_{0}^{\infty}=\frac{\pi}{2b}\cdot \ln(a^2+b^2)$

$So $I =\int_{0}^\infty \frac{\log{x}\,\text{d}x}{(x + a)^2 + b^2} = \frac{\ln(a^2+b^2)\pi}{4b}$

$Now Put $b=a\;,$ We get $I = \int_{0}^\infty \frac{\log{x}\,\text{d}x}{(x + a)^2 + a^2} = \frac{\pi \cdot \ln(2a^2)}{4a}$

juantheron · Mar 2, 2016

Re: MX2 2016 Integration Marathon

$If $I = \int_{0}^{1}(1-x^{50})^{100}dx$ and $J = \int_{0}^{1}(1-x^{50})^{101}dx\;,$ Then $\frac{I}{J} =$

Paradoxica · Mar 2, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$If $I = \int_{0}^{1}(1-x^{50})^{100}dx$ and $I = \int_{0}^{1}(1-x^{50})^{101}dx\;,$ Then $\frac{I}{J} =$

$$\noindent Integrating by parts on $J$ gives $\left[x(1-x^{50})^{101}\right]_0^1 + 101\cdot 50\int_0^1 x^{50} (1-x^{50})^{100}\, $d$x\\$Using Reduction, $101\cdot 50\cdot I - J = 101\cdot 50\cdot J\\\frac{I}{J} = 1+\frac{1}{101\cdot 50}$

Paradoxica · Mar 2, 2016

Re: MX2 2016 Integration Marathon

$\lim_{n \to \infty} \int \frac{\text{d}x}{x\prod_{k=1}^n \cos{\frac{x}{2^k}}}$

KingOfActing · Mar 2, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\lim_{n \to \infty} \int \frac{\text{d}x}{x\prod_{k=1}^n \cos{\frac{x}{2^k}}}$

$\sin{x}= 2\sin{\frac{1}{2}x}\cos{\frac{1}{2}x} = 2^2\sin{\frac{1}{4}x}\cos{\frac{1}{2}x}\cos{\frac{1}{4}x} =\ldots = 2^n\sin{\frac{1}{2^n}x} \prod_{k=1}^n\cos\left({\frac{1}{2^k}x}\right )\\\therefore \prod_{k=1}^\infty\cos\left({\frac{1}{2^k}x}\right ) = \sin{x}\lim_{n\to\infty}\frac{2^{-n}}{\sin{\frac{1}{2^n}x}}\stackrel{u = 2^{-n}}{=} \frac{\sin{x}}{x} \lim_{u\to 0}\frac{ux}{\sin{ux}} = \frac{\sin{x}}{x} \times 1 = \frac{\sin{x}}{x}\\\therefore\lim_{n\to\infty}\int\frac{dx}{x\prod_{k=1}^n\cos\left({\frac{1}{2^k}x}\right )} = \int \frac{dx}{x\times\frac{\sin{x}}{x}} = \int\csc{x}\,dx = -\ln|\csc{x} + \cot{x}| + C$

omegadot · Mar 2, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\lim_{n \to \infty} \int \frac{\text{d}x}{x\prod_{k=1}^n \cos{\frac{x}{2^k}}}$

$\noindent Consider the finite product term first. Note that$

$\begin{align*} \sin x &= 2 \sin \frac{x}{2} \cos \frac{x}{2} = 2^1 \sin \frac{x}{2^1} \cos \frac{x}{2^1}\\&= 2 \cdot 2 \sin \frac{x}{4} \cos \frac{x}{4} \cdot \cos \frac{x}{2} = 2^2 \sin \frac{x}{2^2} \cos \frac{x}{2^2} \cos \frac{x}{2^1}\\& = 2^2 \cdot 2 \sin \frac{x}{8} \cos \frac{x}{8} \cdot \cos \frac{x}{2^2} \cos \frac{x}{2^1} = 2^3 \sin \frac{x}{2^3} \cos \frac{x}{2^3} \cos \frac{x}{2^2} \cos \frac{x}{2^1}\\ &\quad \vdots\\ &= 2^n \sin \frac{x}{2^n} \prod^n_{k = 1} \cos \frac{x}{2^n},\end{align*}$

$$\noindent a result which can be proved by induction on $n \in \mathbb{N}.$ Thus$\\\begin{align*} \prod^n_{k = 1} \cos \frac{x}{2^k} &= \frac{\sin x}{2^n \sin \frac{x}{2^n}}.\end{align*}$

$$\noindent Stepping lightly over issues associated with uniform convergence (which is not part of MX2), the limit can be moved under the integral sign allowing us to rewrite the integral as$\\\begin{align*}\lim_{n \to \infty} \int \frac{\text{d}x}{x\prod_{k=1}^n \cos{\frac{x}{2^k}}} &= \int \frac{\text{d}x}{x \lim_{n \to \infty} \prod_{k=1}^n \cos{\frac{x}{2^k}}}\end{align*}$

$$\noindent For the limit we have$\\\begin{align*} \lim_{n \to \infty} \prod^n_{k = 1} \cos \frac{x}{2^n} &= \sin x \lim_{n \to \infty} \frac{1}{2^n \sin \frac{x}{2^n}}\\ &= \frac{\sin x}{x} \lim_{n \to \infty} \left (\frac{\sin 2^{-n} x}{2^{-n} x} \right )^{-1}\\ &= \frac{\sin x}{x}\end{align*}$

$$\noindent Thus\\\begin{align*}\lim_{n \to \infty} \int \frac{\text{d}x}{x\prod_{k=1}^n \cos{\frac{x}{2^k}}} &= \int \frac{dx}{\sin x} = -\ln |\text{cosec} \, x + \cot x| + \cal{C}.\end{align*}$

omegadot · Mar 2, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $ \int \frac{\sin x}{\sin (x - a) \sin (x - b)} \, dx, \quad a \neq b.$

juantheron · Mar 3, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $ \int \frac{\sin x}{\sin (x - a) \sin (x - b)} \, dx, \quad a \neq b.$

$Put $x-a+x-b = 2t\Rightarrow t=x-\frac{a+b}{2}\;,$ Then $dt=dx$

$So $I = \int\frac{\sin \left(t+A\right)}{\sin(t+B)\cdot \sin(t-B)}dx\;,$ Where $\frac{a+b}{2}=A$ and $\frac{a-b}{2}=B$

$$So $I = \int\frac{\sin t\cdot \cos A+\cos t\cdot \sin A}{\sin^2t-\sin^2B}dt$

$$So $I= \cos A \cdot \int\frac{\sin t}{\cos^2 B-\cos^2 t}dt+\sin A \cdot \int\frac{\cos t}{\sin^2 t-\sin^2 B}dt$

$So $I = \frac{\cos A}{2\cos B}\ln \left|\frac{\cos t-\cos B}{\cos t+\cos B}\right|+\frac{\sin A}{2\sin B}\ln \left|\frac{\sin t-\sin B}{\sin t+\sin B}\right|+\mathcal{C}$

juantheron · Mar 3, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int\frac{1}{(x^4-1)^2}dx$ and $\int_{0}^{1}\frac{x\ln(x)}{\sqrt{1-x^2}}dx$

juantheron · Mar 3, 2016

Re: MX2 2016 Integration Marathon

$Let $\alpha \in \mathbb{R^{+}}$ and $f(\alpha) = \int_{0}^{\infty}\frac{\ln x}{x^2+\alpha x+\alpha^2}dx$

$and $\alpha f(\alpha)-f(1) = \frac{\pi}{\sqrt{3}}\;,$ Then $\alpha$

omegadot · Mar 3, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$$Evaluation of $\int\frac{1}{(x^4-1)^2}dx$

$$\noindent Of course this one can be directly found using a partial fraction decomposition, but to do so is a pretty hard slog. Instead, we will first try and manipulate the integrand a bit.$

$\begin{align*}\int \frac{dx}{(x^4 - 1)^2} &= \frac{1}{4} \int \frac{(-3x^4 + 3) + (3x^4 + 1)}{(x^4 - 1)^2} \, dx\\ &= \frac{1}{4} \int \frac{-3 (x^4 - 1)}{(x^4 - 1)^2} \, dx + \frac{1}{4} \int \frac{3x^4 + 1}{x^2 \left (x^3 - \frac{1}{x} \right )^2} \, dx\\ &= -\frac{3}{4} \int \frac{dx}{x^4 - 1} + \frac{1}{4} \int \frac{3x^2 + \frac{1}{x^2}}{\left (x^3 - \frac{1}{x} \right )^2} \, dx\end{align*}$

$\noindent In the first integral the following partial fraction decomposition can now be used:$\\\begin{align*}\frac{1}{x^4 - 1} = \frac{1}{(x^2 - 1)(x^2 + 1)} = \frac{1/2}{x^2 - 1} - \frac{1/2}{x^2 + 1}\end{align*}\\$ While in the second integral the following substitution is used: $\\\begin{align*}u = x^3 - \frac{1}{x}, du &= \left (3x^2 + \frac{1}{x^2} \right ) \, dx.\end{align*}\\$ Thus$

$\begin{align*}\int \frac{dx}{(x^4 - 1)^2} &= -\frac{3}{8} \int \frac{dx}{x^2 - 1} + \frac{3}{8} \int \frac{dx}{x^2 + 1} + \frac{1}{4} \int \frac{du}{u^2}\\&= - \frac{3}{16} \ln \left | \frac{x - 1}{x + 1} \right | + \frac{3}{8} \tan^{-1} x - \frac{x}{4(x^4 - 1)} + \cal{C}\end{align*}$

HSC 2016 MX2 Integration Marathon (archive) (11 Viewers)

-insert title here-

Well-Known Member

Active Member

Active Member

-insert title here-

Active Member

Active Member

Active Member

-insert title here-

Active Member

Active Member

-insert title here-

-insert title here-

lukewarm mess

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 11)