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

leehuan · Jan 17, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Requires multi-valued gamma function with one real and one imaginary argument, according to Mathematica. So a reduction formula definitely exists.

I'll let it sit there then.

$\int _{ 0 }^{ \pi }{ \frac { \cos ^{ 2 }{ x } }{ 1+\sin { x } } dx }$

Drsoccerball · Jan 17, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I'll let it sit there then.

$\int _{ 0 }^{ \pi }{ \frac { \cos ^{ 2 }{ x } }{ 1+\sin { x } } dx }$

$\int_0^{\pi}{1-\sin{x}}dx$

$I= x+\cos{x}$

$I=\pi -2$

$New Question$

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

omegadot · Jan 18, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$New Question$

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

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

$\noindent Let $u = e^x, du = e^x \, dx$. Thus$

$\begin{align*} \int \frac{\sqrt{e^{2x} - 1}}{e^x + 1} \, dx &= \int \frac{\sqrt{u^2 - 1}}{u^2 + u} \, du \end{align*}$

$\noindent Now let $u = \sec \theta, du = \sec \theta \tan \theta \, d\theta$. So$

$\begin{align*} \int \frac{\sqrt{e^{2x} - 1}}{e^x + 1} \, dx &= \int \frac{\tan^2 \theta}{\sec \theta + 1} \, d\theta\\ &= \int \frac{\sec^2 \theta - 1}{\sec \theta + 1} \, d\theta\\ &= \int (\sec \theta - 1) \, d\theta\\ &= \ln |\sec \theta + \tan \theta| - \theta + \cal{C}\\ &= \ln |u + \sqrt{u^2 - 1}| - \cos^{-1} \left (\frac{1}{u} \right ) + \cal{C}\\ &= \ln |e^x + \sqrt{e^{2x} - 1} | - \cos^{-1} (e^{-x}) + \cal{C}\end{align*}$

$\noindent \textbf{New Question}$

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

leehuan · Jan 18, 2016

Re: MX2 2016 Integration Marathon

$\\ { u }^{ 2 }=x+1+\frac { 1 }{ x } \Rightarrow 2u\, du=\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx\\ \int { \frac { 1-x }{ \left( 1+x \right) \sqrt { x+{ x }^{ 2 }+{ x }^{ 3 } } } dx } \\ =\int { \frac { 1-{ x }^{ 2 } }{ \left( 1+2x+{ x }^{ 2 } \right) \left( x \right) \sqrt { \frac { 1 }{ x } +1+x } } dx } \\ =-\int { \frac { 1-\frac { 1 }{ { x }^{ 2 } } }{ \left( \frac { 1 }{ x } +2+x \right) \sqrt { \frac { 1 }{ x } +1+x } } dx } \\ =-\int { \frac { 2u\, du }{ \left( { u }^{ 2 }+1 \right) u } } \\ =-2\arctan { u } +C\\ =C-2\arctan { \sqrt { \frac { 1 }{ x } +1+x } }$
---------------------

leehuan said:
$\\ $Find a reduction formula for $\int { \cos { { x }^{ \frac { 1 }{ n } } } dx } , n \in \mathbb N$

Paradoxica · Jan 18, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\\ $Find a reduction formula for $\int { \cos { { x }^{ \frac { 1 }{ n } } } dx } , n \in \mathbb N$

Actually, I just realised, nth root of x is irreducible, so there is no reduction formula, only a general form.

Paradoxica said:
$\noindent a) Show that it is always possible to decompose any expression of the form$\\ \frac{p\sin^2x +q\cos^2x +r\sin x\cos x}{a\sin x+b\cos x} \forall \{a,b,p,q,r \} \in \mathbb{R} \\$ into a sum of 3 simpler expressions, subject to restrictions on the denominator.$ \\ $b) Suppose $a^2 +b^2 = c^2$. Evaluate the integral of the above expression.$

glittergal96 · Jan 18, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Actually, I just realised, nth root of x is irreducible, so there is no reduction formula, only a general form.

$\int \cos(x^{1/n})\, dx = n\int y^{n-1}\cos(y)\, dy$

where $y^n=x$ .

(Consider only the non-negative reals to avoid having to talk about multivalued things.)

This latter integral clearly gives us an integration by parts reduction formula (dropping exponent by two each time). Then it's just a matter of changing the variables back to x to get a reduction formula for the original integral.

Someone else can fill in the details if they like.

Paradoxica · Jan 18, 2016

Re: MX2 2016 Integration Marathon

glittergal96 said:
$\int \cos(x^{1/n})\, dx = n\int y^{n-1}\cos(y)\, dy$

where $y^n=x$ .

(Consider only the non-negative reals to avoid having to talk about multivalued things.)

This latter integral clearly gives us an integration by parts reduction formula (dropping exponent by two each time). Then it's just a matter of changing the variables back to x to get a reduction formula for the original integral.

Someone else can fill in the details if they like.

Yes, but in order for a reduction formula to exist, we have to make the "n" which the y-substitution depends on independent to the "n" which the reduction formula depends on.

glittergal96 · Jan 18, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Yes, but in order for a reduction formula to exist, we have to make the "n" which the y-substitution depends on independent to the "n" which the reduction formula depends on.

Ah of course, silly me.

.

KX · Jan 18, 2016

Re: MX2 2016 Integration Marathon

lol why are you using your alt?

seanieg89 · Jan 18, 2016

Re: MX2 2016 Integration Marathon

KX said:
lol why are you using your alt?

Was in my office and knew my alt password off the top of my head, but would have needed to check one of those password reset emails to get my password for the main acct.

Drsoccerball · Jan 18, 2016

Re: MX2 2016 Integration Marathon

#report for boosting.

omegadot · Jan 19, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{x^4}{1 + \tan x + \sec x} \, dx.$

Paradoxica · Jan 19, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Evaluate $\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{x^4}{1 + \tan x + \sec x} \, dx.$

$$\noindent$ I = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{x^4}{1 + \tan x + \sec x} $d$x = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{x^4}{1 - \tan x + \sec x} $d$x $ reflection property$ \\ 2I = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} x^4 \frac{2+2\sec{x}}{(1+\sec{x})^2 - \tan^2{x}}$d$x = \int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} x^4 \frac{2+2\sec{x}}{2+2\sec{x}}$d$x = 2\int_0^{\frac{\pi}{4}} x^4 $d$x \\ I = \frac{\pi^5}{5120}$

Paradoxica said:
$\noindent a) Show that it is always possible to decompose any expression of the form$\\ \frac{p\sin^2x +q\cos^2x +r\sin x\cos x}{a\sin x+b\cos x} \forall \{a,b,p,q,r \} \in \mathbb{R} $ (A.K.A. use partial fractions)$ \\$ into a sum of 3 simpler expressions, subject to restrictions on the denominator.$ \\ $b) Suppose $a^2 +b^2 = c^2$. Evaluate the integral of the above expression.$

Drsoccerball · Jan 19, 2016

Re: MX2 2016 Integration Marathon

$New Question$

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

Drsoccerball · Jan 21, 2016

Re: MX2 2016 Integration Marathon

$Here's another one I enjoyed.$

$\int_0^{1}{\frac{x^4(1-x)^4}{1+x^2}}dx$

leehuan · Jan 21, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$Here's another one I enjoyed.$

$\int_0^{1}{\frac{x^4(1-x)^4}{1+x^2}}dx$

Show that 22/7 > pi

Drsoccerball · Jan 21, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Show that 22/7 > pi

The value of the final integral becomes :

$I= \frac{22}{7} -\pi$

$Since the graph is greater than 0 between the domain : $ 0 \leq x \leq 1 $ This means that the area is greater than 0$

$This is due to the fact that the only x intercept is $ x=1 $ and the graph is always positive due to the even powers.$

$\therefore \frac{22}{7} -\pi > 0$

$\therefore \frac{22}{7} > \pi$

Paradoxica · Jan 21, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$New Question$

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

$$\noindent$ x = \sin{(\theta -\frac{\pi}{4})} \Leftrightarrow \; $d$x = \cos{(\theta -\frac{\pi}{4})}$d$\theta \\ I = \frac{1}{2\sqrt{2}}\int \frac{\cos \theta + \sin \theta}{\sin^2 \theta}$d$\theta = -\frac{1}{2\sqrt{2}}\left (\csc \theta + \log (\csc \theta + \cot \theta ) \right )\\ -2I = \frac{1}{(\sin(\theta - \frac{\pi}{4}) + \cos(\theta -\frac{\pi}{4}))}+\frac{1}{\sqrt{2}}\log \left(\frac{\sqrt{2} + \cos(\theta -\frac{\pi}{4}) - \sin(\theta -\frac{\pi}{4})}{\sin(\theta -\frac{\pi}{4}) +\cos(\theta -\frac{\pi}{4})} \right) \\ \int{\frac{1}{(x+\sqrt{1-x^2})^2}}$d$x = \frac{-\frac{1}{2}}{x+\sqrt{1-x^2}}-\frac{1}{2\sqrt{2}}\log \left(\frac{\sqrt{2}+\sqrt{1-x^2}-x}{x+\sqrt{1-x^2}} \right) +C$

Drsoccerball said:
$Here's another one I enjoyed.$

$\int_0^{1}{\frac{x^4(1-x)^4}{1+x^2}}dx$

ahahahaha... how cute

$$\noindent Evaluate $\int_{0}^{1} \frac{8x^5 (1-x)^2 (324889-120736x^2)}{445625(1+x^2)}\log^2{x} \; $d$x\\$From this, prove $\pi^3 >31$

omegadot · Jan 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Evaluate $\int_{0}^{1} \frac{8x^5 (1-x)^2 (324889-120736x^2)}{445625(1+x^2)}\log^2{x} \; $d$x\\$From this, prove $\pi^3 >31$

$$\noindent Paradoxica, I think this question should really be in the Extracurricular Integration Marathon, but here goes anyway.$

$$\noindent Decomposing the rational part of the integrand into partial fractions, while in principal is not hard, in this particular case is extremely tedious. On doing so we have$\\\begin{align*}\frac{8x^5 (1 - x)^2 (324\,889 - 120\,736 x^2)}{445\,625 (1 + x^2)} &= -\frac{965\,888}{445\,625} x^7 + \frac{1\,931\,776}{445\,625} x^6 + \frac{2\,599\,112}{445\,625} x^5 - 16 x^4 + 16 x^2 - 16 + \frac{16}{x^2 + 1}\end{align*}$

$\noindent From the decomposition we see in the integrand we have a number of terms of the form $x^n \ln^2 x$ that need to be integrated.$

$\noindent Consider $I_n = \int x^n \ln^2 x \, dx$, where $n$ is a non-negative integer. On applying IBP twice it can really be shown that$

$I_n = \frac{x^{n + 1}}{(n + 1)^3} \left [(n + 1)^2 \ln^2 x - 2(n + 1) \ln x + 2 \right ].$

$$\noindent On applying the limits of integration one has$\\I_n = \int^1_0 x^n \ln^2 x \, dx = \frac{2}{(n + 1)^3}.$

$$\noindent Here the following two limits have been used:$\\\begin{align*}\lim_{x \rightarrow 0^+} x^n \ln x = 0 \quad \mbox{and} \quad \lim_{x \rightarrow 0^+} x^n \ln^2 x = 0, \,\,\, \mbox{for} \,\, n \geqslant 1.\end{align*}$

$\noindent From the final term in the partial fraction decomposition we are presented with the integral $\int^1_0 \frac{\ln^2 x}{1 + x^2} \, dx.$ No elementary form for this integral exists by it is still possible to find its value.$

$$\noindent If we start with the following geometric series$\\\begin{align*}\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \cdots = \sum_{k \geqslant 0} x^k,\end{align*}\\$which converges for $|x| < 1$ and replace all appearances of $x$ with $-x^2$ one has$\\\begin{align*}\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots = \sum_{k \geqslant 0} (-1)^k x^{2k},\end{align*}\\$which also converges for $|x| < 1.$ The integral can therefore be rewritten as$

$\begin{align*}\int^1_0 \frac{\ln^2 x}{1+ x^2} \, dx &= \int^1_0 \ln^2 x \sum_{k \geqslant 0} (-1)^k x^{2k} \, dx\\&= \sum_{k \geqslant 0} (-1)^k \int^1_0 x^{2n} \ln^2 x \, dx\end{align*}$

$$\noindent But we have already found integrals of this form. From our recurrence relation $I_n$, if $n$ is replaced with $2k$ one has$\\\int^1_0 x^{2k} \ln^2 x \, dx = \sum_{k \geqslant 0} (-1)^k \frac{2}{(2k + 1)^3} = 2 \sum_{k \geqslant 0} \frac{(-1)^k}{(2k + 1)^3}.$

$$\noindent Now it turns out the sum is a rather famous sum and has a simple closed form (for those in the know our sum is equal to the \textit{Dirichlet beta function} $\beta (s)$ with $s = 3$). Thus$\\\begin{align*}\int^1_0 \frac{\ln^2 x}{1 + x^2} \, dx &= 2 \cdot \beta (3) = 2 \cdot \frac{\pi^3}{32} = \frac{\pi^2}{16}.\end{align*}$

$$\noindent On putting all the pieces together one has$\\\begin{align*}\int_{0}^{1} \frac{8x^5 (1-x)^2 (324889-120736x^2)}{445625(1+x^2)}\ln^2{x} \, dx &= -\frac{965 \,888}{445\,625} I_7 + \frac{1\,931\,776}{445\,625} I_6 + \frac{2\,599\,112}{445\,625} I_5 - 16 I_4 + 16 I_2 - 16 I_0 + 16 \cdot \frac{\pi^3}{16}\\&= -\frac{965\,888}{445\,625} \cdot \frac{2}{8^3} + \frac{1\,931\,776}{445\,625} \cdot \frac{2}{7^3} + \frac{2\,599\,112}{445\,625} \cdot \frac{2}{6^3} - 16 \cdot \frac{2}{5^3} + 16 \cdot \frac{2}{3^3} - 16 \cdot \frac{2}{1^3} + \pi^3\\&= \pi^3 - 31\end{align*}$

$$\noindent For the required inequality, note that all terms appearing in the integrand, bar one, are positive for all $x$. The term $(324\,889 - 120\,736x^2)$, which geometrically is an inverted parabola, is however positive on the interval $(0,1)$. So since the integrand is positive for all $x \in (0,1)$ we have$\\\begin{align*}\int_{0}^{1} \frac{8x^5 (1-x)^2 (324889-120736x^2)}{445625(1+x^2)}\ln^2{x} \, dx &= \pi^3 - 31 > 0\\\Rightarrow \pi^3 &> 31,\end{align*}\\$as required to show. \fbox{Wow!!}$

omegadot · Jan 22, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

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

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