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

Paradoxica · Mar 3, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$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 $\alpha$

$$\noindent Substitute $\frac{a^2}{y} = x$, reduce and integrate as an arctangent. Then $\alpha = e\sqrt{e}$

Paradoxica · Mar 4, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$\int_{0}^{1}\frac{x\ln(x)}{\sqrt{1-x^2}}dx$$

$$\noindent Integrate by parts to fully eliminate the logarithm.$ \\ \int \frac{x\log{x}}{\sqrt{1-x^2}}\,$d$x=-\sqrt{1-x^2}\log{x}+\int \frac{1-x^2}{x\sqrt{1-x^2}}\,$d$x=(1-\log{x})\sqrt{1-x^2}+\int \frac{1-\sqrt{1-x^2}+\sqrt{1-x^2}}{x\sqrt{1-x^2}}\,$d$x=(1-\log{x})\sqrt{1-x^2}+\log{x}+\int\frac{\frac{x}{\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\,$d$x=\sqrt{1-x^2}+(1-\sqrt{1-x^2})\log{x}-\log{(1+\sqrt{1-x^2})}\\$At this point, direct substitution of the borders will yield an indeterminate expression. Transform as follows:$\\\sqrt{1-x^2}+\frac{x^2\log{x}}{1+\sqrt{1-x^2}}-\log{(1+\sqrt{1-x^2})}\\$Direct substitution will not yield an indeterminate form, as we know $x\log{x}$ vanishes in the limit as $x$ vanishes.$\\$The final answer is $\log{2}-1$

Paradoxica · Mar 4, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{2x(x^4+1)^2}{x^{12}+1}\,$d$x$

Drsoccerball · Mar 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{2x(x^4+1)^2}{x^{12}+1}\,$d$x$

$u =x^2 $ Then simplify the expression and use the symmetric substitution $ u= \frac{1}{t}+t$

Paradoxica · Mar 4, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{x^4+1}{x^6-1}\,$d$x$

juantheron · Mar 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{x^4+1}{x^6-1}\,$d$x$

$Let $I = \int\frac{x^4+1+x^2-x^2}{x^6-1} = \int\frac{x^4+1+x^2}{x^6-1}dx-\int\frac{x^2}{x^6-1}dx$

$Let $I = \int\frac{1}{x^2-1}dx-\frac{1}{3}\int\frac{(x^3)^{'}}{(x^3)^2-1}dx$

$So $I = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|-\frac{1}{3}\cdot \frac{1}{2}\cdot \ln \left|\frac{x^3-1}{x^3+1}\right|+\mathcal{C}$

juantheron · Mar 4, 2016

Re: MX2 2016 Integration Marathon

$If $a,b>0\;,$ Then $\int_{0}^{\pi}\frac{\sin^2 x}{a^2-2ab\cos x+b^2}dx\;,$

$and value of above integral if $b>a>0$

juantheron · Mar 4, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int_{0}^{1}\left(tx+1-x\right)^ndx\;,$ Where $n\in \mathbb{Z^{+}}$

$and $t$ is a parameter independent of $x$

$Hence show that $\int_{0}^{1}x^k\left(1-x\right)^{n-k}dx = \left[\binom{n}{k}(n+1)\right]^{-1}$ for $k=0,1,2,3,.....,n$

juantheron · Mar 4, 2016

Re: MX2 2016 Integration Marathon

$Let $I = \int\frac{1}{(x^4-1)^2}dx = \frac{1}{4}\int \left[\frac{1}{x^3}\cdot \frac{4x^3}{(x^4-1)^2}\right]dx$

$Using Integration by parts, We get$

$So $I = -\frac{1}{4}\cdot \frac{1}{x^4-1}-\frac{3}{4}\int\frac{1}{x^4\cdot (x^4-1)}dx$

$So $I = -\frac{1}{4}\cdot \frac{1}{x^4-1}-\frac{3}{4}\int \left[\frac{1}{x^4-1}-\frac{1}{x^4}\right]dx$

$So $I = -\frac{1}{4}\cdot \frac{1}{x^4-1}-\frac{3}{8}\int \left[\frac{(x^2+1)-(x^2-1)}{x^4-1}-\frac{1}{x^4}\right]dx$

Paradoxica · Mar 4, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$If $a,b>0\;,$ Then $\int_{0}^{\pi}\frac{\sin^2 x}{a^2-2ab\cos x+b^2}dx\;,$

$and value of above integral if $b>a>0$

$$\noindent$=\int_0^\pi \frac{1-\cos^2{x}}{a^2+b^2-2ab\cos{x}}\,$d$x=\frac{1}{4a^2b^2}\int_0^\pi \frac{(a^2+b^2)^2-4a^2b^2\cos^2{x}-(a^2-b^2)^2}{a^2+b^2-2ab\cos{x}}\,$d$x\\=\int_0^\pi \frac{a^2+b^2}{4a^2b^2}+\frac{\cos{x}}{2ab}-\frac{(a^2-b^2)^2}{a^2+b^2-2ab\cos{x}}\,$d$x\\$Using the Weierstrass Substitution on the final part, the integral evaluates to $\frac{\pi}{2b^2}$

Note: Why do I have a different answer if I instead convert the denominator into a difference of two squares?

Carrotsticks · Mar 5, 2016

Re: MX2 2016 Integration Marathon

Test

http://i0.kym-cdn.com/photos/images/original/000/002/116/Longcat_War.jpg

Paradoxica · Mar 5, 2016

Help i'm trapped in the title and can only speak once every 2 years

$$\noindent$ \int 20n x^{5n-1} \tan^{-1}\left( \frac{(x^n-1)x^n(x^n+1)}{(x^{2n}-\sqrt{3}x^n+1)(x^{2n}+\sqrt{3}x^n+1)} \right)\,$d$x$

$$\noindent$ \int \tan^{-1}(2^{2n-1}x^2)\,$d$x$

Carrotsticks · Mar 5, 2016

Re: MX2 2016 Integration Marathon

http://i0.kym-cdn.com/photos/images/original/000/002/116/Longcat_War.jpg

Drsoccerball · Mar 6, 2016

I will get you out.

$\int{5^x5^{5^x}5^{5^5^{x}}}dx$

Paradoxica · Mar 6, 2016

Save meeeee!!!!

Drsoccerball said:
$\int{5^x5^{5^x}5^{5^{5^x}}}\,$d$x$

$$\noindent By inspection, the $x$ derivative of $5^{5^{5^x}}$ is $(\log^3{5}) 5^x 5^{5^x} 5^{5^{5^x}}\\\int 5^x 5^{5^x} 5^{5^{5^x}}\,$d$x = \frac{5^{5^{5^x}}}{\log^3{5}}+\cal{C}$

BLIT2014 · Mar 7, 2016

Re: MX2 2016 Integration Marathon

leehuan · Mar 8, 2016

Re: MX2 2016 Integration Marathon

$\int{dx}$

Lol ok I'm sorry.

omegadot · Mar 9, 2016

Re: Help i'm trapped in the title and can only speak once every 2 years

Paradoxica said:
$$\noindent$ \int \tan^{-1}(2^{2n-1}x^2)\,$d$x$

$$\noindent Please let me know if there is a simpler way to do this one.$

$$\noindent Let $V = \int \tan^{-1} (2^{2n - 1} x^2) \, dx.$ Now let $2^{2n - 1} x^2 = \tan \theta, 2^{2n} x \, dx = \sec^2 \theta \, d\theta$ which after a little algebra becomes $dx = \frac{\sec^2 \theta}{2^{n + \frac{1}{2}} \sqrt{\tan \theta}} \, d\theta.$

$\begin{align*}V &= \frac{1}{2^{n + \frac{1}{2}}} \int \frac{\theta \sec^2 \theta}{\sqrt{\tan \theta}} \, d\theta\\&= \frac{1}{2^{n + \frac{1}{2}}} \theta \sqrt{\tan \theta} - \frac{1}{2^{n + \frac{1}{2}}} \int \sqrt{\tan \theta} \, d\theta + \frac{1}{2^{n + \frac{3}{2}}} \int \frac{\theta \sec^2 \theta}{\sqrt{\tan \theta}} \, d\theta \quad (\mbox{by parts})\\\Rightarrow V &= \frac{1}{2^{n - \frac{1}{2}}} \theta \sqrt{\tan \theta} -\frac{1}{2^{n - \frac{1}{2}}} \int \sqrt{\tan \theta} \, dx\end{align*}$

$$\noindent The integral $\int \sqrt{\tan x} \, dx$ can be found in a number of different ways. We will use the following approach: Let $I = \int \sqrt{\tan x} \, dx$ and $J = \int \sqrt{\cot x} \, dx.$

$\begin{align*}I + J &= \int \left (\sqrt{\tan x} + \sqrt{\cot x} \right ) \, dx\\&=\int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{2 \sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x - (\cos^2 x + \sin^2 x)}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin^2 x - 2 \sin x \cos x + \cos^2 x)}} \, dx\\&= \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx\\&= \sqrt{2} \sin^{-1} (\sin x - \cos x) + {\cal{C}}_1\quad (*)\end{align*}$

$$\noindent And\\\begin{align*}I - J &= \int \left (\sqrt{\tan x} - \sqrt{\cot x} \right ) \, dx\\&= \int \frac{\sin x - \cos x}{\sqrt{\sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{2 \sin x \cos x}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{(\cos^2 x + \sin^2 x) + 2\sin x \cos x - 1}} \, dx\\&= \sqrt{2} \int \frac{\sin x - \cos x}{\sqrt{(\cos x + \sin x)^2 - 1}} \, dx\\&= -\sqrt{2} \ln \left |\sqrt{(\cos x + \sin x)^2 - 1} + \sin x + \cos x \right | + {\cal{C}}_2\\ &= -\sqrt{2} \ln \left |\sqrt{\sin 2x} + \sin x + \cos x \right | + {\cal{C}}_2 \quad (**)\end{align*}$

$\noindent Adding$ (*)$ and $(**)$ gives$

$2 I = \sqrt{2} \sin^{-1} (\sin x - \cos x) - \sqrt{2} \ln \left |\sqrt{\sin 2x} + \sin x + \cos x \right | + {\cal{C}}$

$\noindent or$

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

$$\noindent Thus$\\\begin{align*}V &= \frac{1}{2^{n - \frac{1}{2}}} \theta \sqrt{\tan \theta} - \frac{1}{2^n} \sin^{-1} (\sin \theta - \cos \theta) - \frac{1}{2^n} \ln \left |\sqrt{\sin 2\theta} + \sin \theta + \cos \theta \right | + \cal{C}\end{align*}$

$$\noindent Noting that $\sin \theta = \frac{2^{2n - 1} x^2}{\sqrt{2^{4n - 2} x^4 + 1}}$ and $\cos \theta = \frac{1}{\sqrt{2^{4n - 2} + 1}}$. Substituting into the expression for $V$, after some algebra, we finally have$\\ \begin{align*}\int \tan^{-1} (2^{2n - 1} x^2) \, dx &= x \tan^{-1} (2^{2n - 1} x^2) - \frac{1}{2^n} \sin^{-1} \left (\frac{2^{2n - 1} x^2 - 1}{\sqrt{2^{4n - 2} x^4 + 1}} \right ) + \frac{1}{2^n} \ln \left |\frac{2^{2n - 1} x^2 + 2^n x + 1}{\sqrt{2^{4n - 2} x^4 + 1}} \right | + \cal{C}.\end{align*}$

omegadot · Mar 9, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{Next Question}$

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

Paradoxica · Mar 9, 2016

pi is a rational number!!!!

omegadot said:
-arduous amounts of working out-

$$\noindent The fastest way to integrate this is to consider the evaluation of $\int \tan^{-1}{\frac{y^2}{2}}$d$y $ and then directly substitute $y = 2^n x$

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