haha, i saw that video once, it's a fairly easy integral, so i consider they are trying to prove your intelligence and speed altogether.

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- Jan 27th 2011, 05:48 AMKrizalid
haha, i saw that video once, it's a fairly easy integral, so i consider they are trying to prove your intelligence and speed altogether.

- Jan 27th 2011, 08:28 AMFernandoRevilla

Some books (for example*Demidovich*) include the following formula in the Table for immediate integrals:

$\displaystyle \displaystyle\int \dfrac{dx}{a^2-x^2}=\dfrac{1}{2a}\ln \left |{\dfrac{a+x}{a-x}}\right |+C\quad (a\neq 0)$

In that case, and taking into account that we needn't an elegant development, and that the euclidean division is almost immediate, the integral can be computed in less than 20 seconds. :)

Fernando Revilla - Jan 27th 2011, 11:42 AMtom@ballooncalculus
Nice case of the 'addition/subtraction' method in no. 14.

$\displaystyle \displaystyle{\ \ \frac{1}{x^2(x^4 + 1)^{3/4}}}$

$\displaystyle \displaystyle{=\ \frac{(x^4 + 1) - x^4}{x^2(x^4 + 1)^{3/4}}$

$\displaystyle \displaystyle{=\ \frac{(x^4 + 1)^{3/4}(x^4 + 1)^{1/4} - x^4}{x^2(x^4 + 1)^{3/4}}$

$\displaystyle \displaystyle{=\ \frac{(x^4 + 1)^{1/4}}{x^2}\ -\ \frac{x^2}{(x^4 + 1)^{3/4}}$

As with no.s 2 and 16, start parts with one half of it and it's all over 'too' soon...

http://www.ballooncalculus.org/draw/parts/eight.png

... where (key in spoiler) ...

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jan 27th 2011, 12:28 PMtom@ballooncalculus
Is this party over? Oh well, anyway... no.11

http://www.ballooncalculus.org/draw/parts/nine.png

... where (key in spoiler) ...

__Spoiler__:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote! - Jan 28th 2011, 04:14 AMJester
2) $\displaystyle \displaystyle \int 2 \ln x + \ln^2 x\, dx = \int \left( x \ln^2x \right)'\,dx$

- Jan 28th 2011, 02:16 PMcarrot
$\displaystyle \int \sin[(n+1)x] \sin^{n-1}x {dx} = \int [\sin(nx)\cos x+\cos(nx)\sin x ] \sin^{n-1}x {dx}$

$\displaystyle = \frac{1}{n} \int \sin(nx) d\sin^n x+\int \cos(nx) \sin^n x {dx} =\frac{1}{n} \sin(nx)\sin^n x + c$ - Jan 31st 2011, 01:48 AMTheCoffeeMachine
I thought the following three integrals were very similar to those in the test:

1. $\displaystyle \int\frac{2x\cos{x}+\sin{x}}{4\sqrt{x}\sqrt{\sin{x }\sqrt{x}}}\;{dx}.$

2. $\displaystyle \int \frac{2\sin^{-1}{x}+2\cos^{-1}{x}}{\sqrt{1-x^2}}\;{dx}.$

3. $\displaystyle \int\frac{1}{x+x\log^2{x}}\;{dx}$.

All three can be solved without explicit substitution (the last being the easiest). - Jan 31st 2011, 12:21 PMDrexel28
- Feb 1st 2011, 01:59 AMTheCoffeeMachine
Doh, I didn't notice that! (Doh) I was thinking more along the lines of:

$\displaystyle \begin{aligned} \displaystyle \int \frac{2\sin^{-1}{x}+2\cos^{-1}{x}}{\sqrt{1-x^2}}\;{dx} & = \int \frac{2\sin^{-1}{x}}{\sqrt{1-x^2}}+\frac{2\cos^{-1}{x}}{\sqrt{1-x^2}}\;{dx} \\& = \int 2\sin^{-1}{x}(\sin^{-1}{x})' -2\cos^{-1}{x}(\cos^{-1}{x})'\;{dx} \\& = \int \left[\left(\sin^{-1}{x}\right)^2\right]'-\left[\left(\cos^{-1}{x}\right)^2\right]' \;{dx} \\& = \left(\sin^{-1}{x}\right)^2-\left(\cos^{-1}{x}\right)^2+k. \end{aligned}$