HSC 2016 MX2 Integration Marathon (archive) (9 Viewers)

Drsoccerball · Feb 14, 2016

Re: MX2 2016 Integration Marathon

RachelGreen said:
I've just started integration 4u, and am looking through this read from the beginning. How do you guys do Integration by Parts so quickly? I'm sick and tired of writing u and u' and v' and v

$I'm not sure what you mean but my extension 2 teacher suggested this method:$

$Writing in this order:$
u=... u'=...

v'=... v=...

$You first multiply v diagonally and then vertically but the vertical one becomes a negative integral$

$uv -\int{u'v}$

RachelGreen · Feb 14, 2016

Re: MX2 2016 Integration Marathon

Yeah I learnt it like that as well! Is that the only way to do it, or is there a faster way?

InteGrand · Feb 14, 2016

Re: MX2 2016 Integration Marathon

RachelGreen said:
I've just started integration 4u, and am looking through this read from the beginning. How do you guys do Integration by Parts so quickly? I'm sick and tired of writing u and u' and v' and v

$\noindent You don't need to bother writing $u$ and $v^\prime$ etc. You can just look at the product in the integrand, and pick one of them to be the one to differentiate, and the other to integrate. Then write the IBP as integrate one of them times leaving the other, minus the integral of integrate that same one and differentiate the other. In other words, you basically skip the steps of writing $u$ and $v^\prime$, as well as the step of writing out ``$\int u \text{ d}v = vu - \int v \text{ d}u$''. You just do these steps without writing them out with $u$'s and $v$'s.$

RachelGreen · Feb 14, 2016

Re: MX2 2016 Integration Marathon

OhHH okay thank u

Drsoccerball · Feb 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
New Question:

$\int_0^{\infty}{\frac{x \ln{x}}{(a^2+x^2)(a^2x^2+1)}}dx$

bump.

seanieg89 · Feb 15, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
New Question:

$\int_0^{\infty}{\frac{x \ln{x}}{(a^2+x^2)(a^2x^2+1)}}dx$

$J=\int_0^\infty \frac{x\log{x}}{(x^2+a^2)(a^2x^2+1)}\, dx\\ \\ =\int_\infty^0 \frac{\frac{1}{x}\log{\frac{1}{x}}}{\left(\frac{1}{x^2}+a^2\right)\left(\frac{a^2}{x^2}+1\right)}\, \frac{-dx}{x^2}\\ \\ \\ =-\int_0^\infty \frac{x\log{x}}{(x^2+a^2)(a^2x^2+1)}\, dx\\ \\ =-J \\ \\ \Rightarrow J=0.$

leehuan · Feb 15, 2016

Re: MX2 2016 Integration Marathon

RachelGreen said:
I've just started integration 4u, and am looking through this read from the beginning. How do you guys do Integration by Parts so quickly? I'm sick and tired of writing u and u' and v' and v

InteGrand said:
$\noindent You don't need to bother writing $u$ and $v^\prime$ etc. You can just look at the product in the integrand, and pick one of them to be the one to differentiate, and the other to integrate. Then write the IBP as integrate one of them times leaving the other, minus the integral of integrate that same one and differentiate the other. In other words, you basically skip the steps of writing $u$ and $v^\prime$, as well as the step of writing out ``$\int u \text{ d}v = vu - \int v \text{ d}u$''. You just do these steps without writing them out with $u$'s and $v$'s.$

Precisely.

Nobody said you HAD to write out u and dv. It's the same as the product rule: if you can, USE it immediately

Paradoxica · Feb 15, 2016

Re: MX2 2016 Integration Marathon

$\int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx$

omegadot · Feb 16, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent Integrating by parts gives $\frac{\pi}{4}\log{2}-\int_0^1 \frac{\log{(x+1)}}{x^2+1}$d$x \\ $Which is a standard integral, and that gives a final answer of $\frac{\pi}{8}\log{2} \\ $Comment: Still trying to think of a more direct method of evaluation.$

$$\noindent Here is a more direct method which avoids the $\int^1_0 \frac{\ln (x + 1)}{x^2 + 1} \, dx$ integral.$

$\noindent Let $u = \frac{1 - x}{1 + x} \Rightarrow x = \frac{1 - u}{u + 1}, dx = -\frac{2}{(u + 1)^2} \, du$. For the limits, when $x = 1, u = 0$ and when $x = 0, u = 1$. Thus$

$\begin{align*}\int^1_0 \frac{\tan^{-1} x}{1 + x} \, dx &= \int^1_0 \frac{1}{1 + u} \tan^{-1} \left (\frac{1 - u}{1 + u} \right ) \, du\end{align*}$

$$\noindent Using the result $\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left (\frac{a - b}{1 + ab} \right )$ we have$\\\tan^{-1} 1 - \tan^{-1} u = \frac{\pi}{4} - \tan^{-1} u = \tan^{-1} \left (\frac{1 - u}{1 + u} \right ).$

$$\noindent Thus$\\\begin{align*}\int^1_0 \frac{\tan^{-1} x}{1 + x} \, dx &= \frac{\pi}{4} \int^1_0 \frac{du}{1 + u} - \int^1_0 \frac{\tan^{-1} u}{1 + u} \, du\\\Rightarrow 2 \int^1_0 \frac{\tan^{-1} x}{1 + x} \, dx &= \frac{\pi}{4} \ln 2\\\Rightarrow \int^1_0 \frac{\tan^{-1} x}{1 + x} \, dx &= \frac{\pi}{8} \ln 2\end{align*}$

omegadot · Feb 16, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \left(\frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 dx$

$$\noindent We shall solve this one using what I call the {\em reverse quotient rule}.$

$$\noindent As is well known, from the quotient rule we have $\left (\frac{f}{g} \right )' = \frac{f' g - g' f}{g^2}, \,\, g \neq 0.$

$$\noindent We now try to write the integrand so that it appears in the form of the right side of the quotient rule. When this is done one will have$\\\int \frac{f' g - g' f}{g^2} \, dx = \int \left (\frac{f}{g} \right )' \, dx = \frac{f}{g} + \cal{C}.$

$\noindent For our integral, clearly $g(x) = x^3 - x + 1$. Noting that $\deg (g) = 3$ while for the integrand, $\deg (\mbox{numerator}) = 4$. With the assumption that $f$ is a polynomial we have $\deg (f) = 2$. So we set $f(x) = ax^2 + bx + c$, where $a,b,$ and $c$ are constants to be determined.$

$\begin{align*}f'g - g'f = (2ax + 3)(x^3 - x + 1) - (3x^2 - 1)(ax^2 + bx + c) &= \left (x^2 - 3x + \frac{1}{3} \right )^2\\-ax^4 + (3 - 3b) x^3 + (-3c - a) x^2 + (b - 3) x + 3 + c &= x^4 - 6x^3 + \frac{29}{3} x^2 - 2x + \frac{1}{9}\end{align*}$

$$\noindent On equating equal coefficients for $x$ we have $a = -1, b = 3, c = -\frac{26}{9}.$ Thus$\\f(x) = - x^2 + 3x - \frac{26}{9}$

$$\noindent So$\\\begin{align*} \int \left (\frac{x^2 - 3x + \frac{1}{3}}{x^3 - x + 1} \right )^2 \, dx &= \int \left (\frac{-x^2 + 3x - \frac{26}{9}}{x^3 - x + 1} \right )' \, dx = \frac{-x^2 + 3x - \frac{26}{9}}{x^3 - x + 1} + \cal{C}\end{align*}$

omegadot · Feb 16, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Find $\int\frac{2x + 5}{x(x + 1)(x + 4)(x+5) + 5} \, dx.$

Paradoxica · Feb 16, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Find $\int\frac{2x + 5}{x(x + 1)(x + 4)(x+5) + 5} \, dx.$

$$\noindent Directly expanding the denominator and recollecting terms gives $\\(x^2 + 5x + 2)^2 +1$. Direct substitution or reverse chain rule yields $\\ \tan^{-1}{(x^2 + 5x + 2)} + \cal{C}$

Paradoxica · Feb 16, 2016

Re: MX2 2016 Integration Marathon

$n\int \sin{(n+1)x}\, \sin^{n-1}{x}\, $d$x$

$n\int \cos{(n+1)x}\, \cos^{n-1}{x}\, $d$x$

$n\int \cos{(n+1)x}\, \sin^{n-1}{x}\, $d$x$

$n\int \sin{(n+1)x}\, \cos^{n-1}{x}\, $d$x$

omegadot · Feb 17, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$n\int \sin{(n+1)x}\, \sin^{n-1}{x}\, $d$x$

$n\int \cos{(n+1)x}\, \cos^{n-1}{x}\, $d$x$

$$\noindent (c) $ n\int \cos{(n+1)x}\, \sin^{n-1}{x}\, $d$x$

$n\int \sin{(n+1)x}\, \cos^{n-1}{x}\, $d$x$

$\noindent I'll do one of these. The rest are done in a similar fashion which I'll save for others to try. Consider the third one (c). Noting that$

$n \cos (n + 1) x \sin^{n - 1} x = n \cos nx \cos x \sin^{n - 1} x - n \sin nx \sin^n x = \frac{d}{dx} (\cos nx \sin^n x)$

$\noindent So we have$

$n \int \cos (n + 1)x \sin^{n - 1} x \, dx = \int d (\cos nx \sin^n x) = \cos nx \sin^n x + \cal{C}.$

omegadot · Feb 17, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Evaluate $\int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, dx$

Paradoxica · Feb 17, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Evaluate $\int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, $d$x$

$$\noindent Observe $\tan^{-1}\sqrt{\frac{1-x}{1+x}} + \frac{\sin^{-1}{x}}{2} \equiv \frac{\pi}{4} \\$ in the valid domain of both functions.$ \\ \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, $d$x = \int_{-1}^1 \frac{\pi}{4\sqrt{1-x^2}} - \frac{\sin^{-1}{x}}{2\sqrt{1-x^2}}\,$d$x \\ $Further observe $\frac{\sin^{-1}{x}}{2\sqrt{1-x^2}}$ is odd and $\frac{\pi}{4\sqrt{1-x^2}}$ is even.$ \\ \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, $d$x = \frac{\pi}{2} \int_0^1 $d$\left(\sin^{-1}{x}\right) = \frac{\pi^2}{4}$

$\int \frac{\text{d}x}{\sqrt{x}(1+\sqrt[3]{x})}$

Drsoccerball · Feb 17, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent Observe $\tan^{-1}\sqrt{\frac{1-x}{1+x}} + \frac{\sin^{-1}{x}}{2} \equiv \frac{\pi}{4} \\$ in the valid domain of both functions.$ \\ \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, $d$x = \int_{-1}^1 \frac{\pi}{4\sqrt{1-x^2}} - \frac{\sin^{-1}{x}}{2\sqrt{1-x^2}}\,$d$x \\ $Further observe $\frac{\sin^{-1}{x}}{2\sqrt{1-x^2}}$ is odd and $\frac{\pi}{4\sqrt{1-x^2}}$ is even.$ \\ \int^1_{-1} \frac{1}{\sqrt{1 - x^2}} \tan^{-1} \left (\sqrt{\frac{1 - x}{1 + x}} \right ) \, $d$x = \frac{\pi}{2} \int_0^1 $d$\left(\sin^{-1}{x}\right) = \frac{\pi^2}{4}$

$\int \frac{\text{d}x}{\sqrt{x}(1+\sqrt[3]{x})}$

How can one simply "observe" that fact... It can be done also through $\tan ^2 \theta = \frac{1-x}{1+x}$

Paradoxica · Feb 17, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
How can one simply "observe" that fact... It can be done also through $\tan ^2 \theta = \frac{1-x}{1+x}$

Differentiate the function, re-integrate it in simpler form, then find the arbitrary constant.

Alternatively, use the tangent double angle formula on the argument.

Recall I do many things by inspection. This is one but many, many of them.

InteGrand · Feb 17, 2016

Re: MX2 2016 Integration Marathon

$\noindent Another way to do that integral would be to split it up between $-1$ to $0$ and $0$ to $1$, use a substitution for the one from $-1$ to $0$ of $u=-x$, and make use of the identity $\tan^{-1} \frac{1}{t}=\frac{\pi}{2}-\tan^{-1} t$ for $t>0$. This allows some integrals to cancel and leave us with $I=\int_0 ^1 \frac{\pi}{2}\frac{1}{\sqrt{1-u^2}}\text{ d}u = \frac{\pi^2}{4}$.$

omegadot · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Alternatively a substitution of $x = \sin \theta$ followed by recalling that $\\\begin{align*}\sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} &= \tan \left (\frac{\theta}{2} + \frac{\pi}{4} \right )\end{align*}\\$ would work as well, though perhaps being expected to simply \textit{recall} that last trig identity may be a bit much.$

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