A Summer of Maths (ASoM) 2016

Original post by Zacken

How so? Neither one of those has a nice anti-derivative.

Duno just running some **** through my head.

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Reply 101

drandy76

13

In a moment of weakness nearly ran it through wolfram alpha, almost

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Reply 102

Original post by Zacken

How so? Neither one of those has a nice anti-derivative.

Easy zain. Took few minutes once i started writing. Pi/2(ln2)
Stop distracting me from d2 revision.
Need to pass otherwise im gna B in AFM.

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Reply 103

Imperion

17

Groups looks so bloody interesting *o*

Reply 104

Student403

19

Original post by drandy76

Unless I've goofed you end up were you started

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Considering it's been posted on this thread by that person, I wouldn't expect anything less :tongue:

Reply 105

drandy76

13

Original post by Student403

Considering it's been posted on this thread by that person, I wouldn't expect anything less :tongue:

I have two ideas and I'm too lazy to attempt either :frown:

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Reply 106

Original post by drandy76

I have two ideas and I'm too lazy to attempt either :frown:

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Na the obvious parts direction is roght. Just carry on going and note some symmetry.
A graph is symmetric about a if
F(2a-x)=F(x) or some simple symmetry of sin which is the same lol.

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Reply 107

Will post sol if any mone wants it. Just dnt wana ruin it.

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Reply 108

OP

Original post by Snasher350

Substituting U=Cos x ? Should end up with a by parts integral with u and arccosU

Doesn't sound like it'll work.

Original post by physicsmaths

Will post sol if any mone wants it. Just dnt wana ruin it.

Did you basically end up finding integral of ln (sin x) between 0 and pi/2? If so, that wasn't really the intended method. :tongue:

Reply 109

drandy76

13

Original post by physicsmaths

Na the obvious parts direction is roght. Just carry on going and note some symmetry.
A graph is symmetric about a if
F(2a-x)=F(x) or some simple symmetry of sin which is the same lol.
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ah i see, was considering taylor expansions lol

Reply 110

Original post by Zacken

Doesn't sound like it'll work.

Did you basically end up finding integral of ln (sin x) between 0 and pi/2? If so, that wasn't really the intended method. :tongue:

Yeh. Not indefinite tho. Just added some **** and pi/2-x and u-pi/2 subs.
Here it is don't look anyone who hasn't solved yet, it is a very nice problem.
ImageUploadedByStudent Room1467150054.370639.jpg

ImageUploadedByStudent Room1467150054.370639.jpg

Was gna do a tanx method etc. Seems easier thenz

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Reply 111

OP

FWIW, my approach:

Unparseable latex formula:

\displaystyle [br]\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}

Now taking

I(0) = 0

gives us

\mathcal{C}

and

\alpha = 1

recovers

\int_0^{\frac{\pi}{2}} x \cot x \, \mathrm{d}x = \frac{\pi}{2} \log 2

.

Reply 112

tridianprime

2

Original post by Insight314

So many people got the Beardon, and I don't. I am furious! :angry:

I will have it in two days though. :h:

I just had a look at the contents page on Amazon.

*goes to order the book*

Reply 113

Original post by Zacken

FWIW, my approach:

Unparseable latex formula:

\displaystyle [br]\begin{align*} I(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\arctan (\alpha \tan x)}{\tan x} \, \mathrm{d}x & \Rightarrow I'(\alpha) = \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{(\alpha \tan x)^2 + 1} \\ & \Rightarrow I'(\alpha) = \frac{\pi}{2(\alpha +1)} \\ & \Rightarrow I(\alpha) = \frac{\pi}{2}\ln (\alpha + 1) + \mathcal{C} \end{align*}

Now taking

I(0) = 0

gives us

\mathcal{C}

and

\alpha = 1

recovers

\int_0^{\frac{\pi}{2}} x \cot x \, \mathrm{d}x = \frac{\pi}{2} \log 2

.

Ah ok. Very nyc.
Shudda thought this was undergrad type ****.

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Reply 114

OP

Pesky people:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \int_0^1 \frac{x^2 - 1}{\ln x} \, \mathrm{d}x \end{equation*}

Reply 115

wlog54

Original post by Zacken

Pesky people:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \int_0^1 \frac{x^2 - 1}{\ln x} \, \mathrm{d}x \end{equation*}

I got ln3 by considering I(a) = integral from 0 to 1 of e^(alnx)/lnx

Reply 116

OP

Original post by wlog54

I got ln3 by considering I(a) = integral from 0 to 1 of e^(alnx)/lnx

The answer is correct, but I'm not entirely sure about your parametrisation. What

a

do you need to recover the original integral? The only way I can see you move on from your parametrisation is by integrating the power series term by term, which isn't exactly savoury.

I'd use something like

\displaystyle I(\alpha) = \int_0^1 \frac{x^{\alpha} - 1}{\ln x}

.

(edited 7 years ago)

Reply 117

ThatPerson