The Proof is Trivial!

$\int^\frac{1}{2}_0 (arcsinx)^2 dx = \frac{\pi^2}{72}-\int^\frac{1}{2}_0\frac{2xarcsinx}{\sqrt{1-x^2}} dx = \frac{\pi^2}{72}+\frac{\pi\sqrt{3}}{6} - 1$

God typing LaTeX on a mobile is tedious.

\Rightarrow I = 4\displaystyle \int \dfrac{u}{(u+1)^{10}} \ dx

Nice job! Though, given how straightforward a lot of these integrals are, I believe the aim is to come up with with most elegant/quick method. I believe that the only chance someone is going to have to complete one of these questions every 50 seconds is to find such elegant methods! Heres mine

Solution 187 (2)

Let $u=\sqrt[4]{x}$,

$\displaystyle \Rightarrow \int \dfrac{1}{\sqrt{x} (\sqrt[4]{x}+1)^{10}} dx = \int \dfrac{(u+1)-1}{(u+1)^{10}} \ dx = \int \dfrac{1}{(u+1)^9} -\dfrac{1}{(u+1)^{10}} du [br]\displaystyle =-\frac{1}{8} (u+1)^{-8} +\frac{1}{9} (u+1)^{-9} +\mathcal{C} = - \dfrac{1}{18}\left( \dfrac{9\sqrt[4]{x}+1}{(\sqrt[4]{x}+1)^{9}}\right)+ \mathcal{C}$

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Problem 184*

Evaluate $\displaystyle \int \frac{x^4}{1-x^2} dx$

Problem 191*


Determine all real numbers $z$ which satisfy this equation, and remarking as to why some roots are disregarded.


$\sqrt{\sqrt{3-z}-\sqrt{z+1}}>\frac{1}{2}$

Solution 184

Let $x = tanhy$

$dx = sech^2y \ dy$

$\Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy$

$= \int tanh^4y \ dy$

$= \int (1-sech^2y)tanh^2y \ dy$

$= \int 1-sech^2y - sech^2ytanh^2y \ dy$

$= y - tanhy - \frac{1}{3} tanh^3y + c$

$= artanhx - x -\frac{1}{3}x^3 + c$

The difference is marginal, I just showed more working, though, you're right, that would have been slightly neater.

Solution 184

Let $x = tanhy$

$dx = sech^2y \ dy$

$\Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy$

$= \int tanh^4y \ dy$

$= \int (1-sech^2y)tanh^2y \ dy$

$= \int 1-sech^2y - sech^2ytanh^2y \ dy$

$= y - tanhy - \frac{1}{3} tanh^3y + c$

$= artanhx - x -\frac{1}{3}x^3 + c$

Solution 184

Let $x = tanhy$

$dx = sech^2y \ dy$

$\Rightarrow I = \int \dfrac{tanh^4ysech^2y}{sech^2y} \ dy$

$= \int tanh^4y \ dy$

$= \int (1-sech^2y)tanh^2y \ dy$

$= \int 1-sech^2y - sech^2ytanh^2y \ dy$

$= y - tanhy - \frac{1}{3} tanh^3y + c$

$= artanhx - x -\frac{1}{3}x^3 + c$

Note $\sqrt{ \sqrt{3-z} - \sqrt{z+1}}$ is only defined for $-1 \leq z < 1$

$\Rightarrow 1 \leq z < 1 + \dfrac{\sqrt{127}}{32}$

*coughcough*

Solution 191

Note $\sqrt{ \sqrt{3-z} - \sqrt{z+1}}$ is only defined for $-1 \leq z < 1$

$\sqrt{ \sqrt{3-z} - \sqrt{z + 1}} > \dfrac{1}{2} \Rightarrow \sqrt{3-z} \cdot \sqrt{z+1} > \dfrac{63}{32}$

$\Rightarrow 1 \leq z < 1 - \dfrac{\sqrt{127}}{32}$

Is there a more concise solution than my page and half of arithmetic?

Typo typo typo

Really?

Of course. I bet you're just one of those reckless bastards who loves introducing additional solutions into equations. Something like this:

http://i.qkme.me/3otjtl.jpg

By the way, the second question is much more difficult. I can't think of anything simpler than L-functions.

Typo


I like to live dangerously

#CARPEDIEM

You love your hyperbolic functions!

A correct and perfectly acceptable solution. However, this was in the head-to-head and there is no chance you could have done that in less than 30 seconds let alone doing most of the working in your head!

My approach on the other hand... (hopefully now my point will be made)

Solution 184 (2)

$\displaystyle \int \frac{x^4}{1-x^2} dx =\int \frac{1-(1-x^2)-x^2(1-x^2)}{1-x^2} dx = artanh(x)-x-\frac{1}{3} x^3 +\mathcal{C}$