ETF2L – Latest activity in “Can you integrate this?” https://staging.etf2l.org/forum/off-topic/topic-28808/ Reply by Theoz https://staging.etf2l.org/forum/off-topic/topic-28808/page-3/?recent=498011#post=498011 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=498011 Wed, 28 May 2014 14:36:18 +0200 Reply by Iller https://staging.etf2l.org/forum/off-topic/topic-28808/page-3/?recent=498011#post=498007 Quoted from Steve!
3.14 is an estimate of pi, so it won't give you the exact result. FriendlyCow — no, they don’t :D sin3x can be integrated to -(1/3)cos3x +c and the x can still be either in radians or degrees, whichever suits you better And as a triple post cause i’m dumb (feel free to merge them anyone, sorry): you can convert your pi into degrees by multiplying by 180 and dividing by pi, hence giving you the 90 degrees Permzilla mentioned
Things can easily get a little confused here, though. Lets take the sine function. We have a degree version and a radian version of the function, where Sin_deg(v) is equal to (or can de defined as) Sin_rad(v*pi/180). Sin_rad(w) is defined the usual way as the y-coordinate on the unit circle for the point corresponding to the angle w (measured in radians). When taking the derivative d/dv it also matters what dv we are talking about, i.e if it is the derivative with respect to the angle measured in degrees or radians. Let v be the angle variable measured in degrees and let w be the variable measured in radians. Then v = w*(180/pi) and dv = (180/pi)dw. If everything is expressed in radians we have (d/dw)sin_rad(w) = cos_rad(w) as the standard formulas tells us. If the argument w is switched to v and measured in degrees we still get (d/dv)sin_rad(v) = cos_rad(v), since the d/dw terms changed to d/dv at the same time the function argument changed from w to v. No extra constants here. However, if everything is expressed in degrees and we also use the degree version of the sine function, we get (d/dv)sin_deg(v) = (by definition) (d/dv)sin_rad(v*(pi/180)) = (pi/180)* cos_rad(v*(pi/180)) = (pi/180)*cos_deg(v). So if degrees are used everywhere including using the degree version of the sine function, we would get extra multiplicative constants, like in the integral of the original poster. That's one of the reason to switch to using radians everywhere instead. No extra constants pops up if it is assumed that the radian version trigonometric functions are used everywhere. In that case the argument itself can be expressed in either radians or degrees.]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=498007 Wed, 28 May 2014 12:54:27 +0200 Reply by BenBazinga https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497802 Quoted from Rex
[...] The whole million? Surely just take my proof and give me $50k. ;)
How sweet!]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497802 Sun, 25 May 2014 18:21:41 +0200 Reply by Rex McPwn https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497801 Quoted from BenBazinga
Got more work for you, Timon! If you proof that the real part of every non-trivial zero of the Riemann zeta function is 1/2, I'll give you 1 000 000$ / 0.24$/ref = 4 166 666 ref and 2 reclaimed for your effort.
The whole million? Surely just take my proof and give me $50k. ;)]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497801 Sun, 25 May 2014 18:18:55 +0200 Reply by BenBazinga https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497800 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497800 Sun, 25 May 2014 18:17:28 +0200 Reply by Rex McPwn https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497799 Quoted from Steve!
Let's entertain that thought: If the question was asking for any other constant, such as sin(3.14/2) in degrees * cos(x), you'd just treat it as a constant and use the product rule for integrating : uv - integration(v*(du/dx)). The answer would therefore be Sin(1.57)*Sin(x) +sin(x)*0 = sinx * sin(1.57) So, treating it as degrees, to 3s.f., your integration would be 0.0274sin(x) +c
It's called integration by parts. And you don't need it since you really should know that integral(acos(x)) = asin(x) + c because you're dealing with a constant, not another function of x.]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497799 Sun, 25 May 2014 18:15:42 +0200 Reply by Tseini94 https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497798 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497798 Sun, 25 May 2014 18:11:34 +0200 Reply by schabdaniel https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497797 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497797 Sun, 25 May 2014 18:06:50 +0200 Reply by Rex McPwn https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497796 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497796 Sun, 25 May 2014 18:05:02 +0200 Reply by joesefff https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497795 Quoted from Rex
[...] You don't but it's a pretty reasonable assumption they expect you to work in radians since sin(pi/2) = 1. They would accept an answer in degrees it just wouldn't be as "neat". From here on up you pretty much want to ditch degrees in favour of radians anyway. Get used to them!
Alrighty. Huge thanks to everyone who bothered to answer!]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497795 Sun, 25 May 2014 18:01:37 +0200 Reply by Rex McPwn https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497794 Quoted from jx53
[...] I have, but lets make an imaginary situation where the value is 3,14/2. No radians. Could that be integrated? If so what would the answer be? How do i know for sure if the question means radians?
You don't but it's a pretty reasonable assumption they expect you to work in radians since sin(pi/2) = 1. They would accept an answer in degrees it just wouldn't be as "neat". From here on up you pretty much want to ditch degrees in favour of radians anyway. Get used to them!]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497794 Sun, 25 May 2014 17:58:33 +0200 Reply by joesefff https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497793 Quoted from Rex
[...] Have you met radians before?
I have, but lets make an imaginary situation where the value is 3,14/2. No radians. Could that be integrated? If so what would the answer be? How do i know for sure if the question means radians?]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497793 Sun, 25 May 2014 17:58:16 +0200 Reply by schabdaniel https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497790 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497790 Sun, 25 May 2014 17:50:01 +0200 Reply by Permzilla https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497788 Quoted from jx53
Yeah. But what if it is sin (3.14/2)
that will work if your calculator is in radians, but most calculators are in degrees by default.]]> generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497788 Sun, 25 May 2014 17:48:41 +0200 Reply by thefriendlycow2 https://staging.etf2l.org/forum/off-topic/topic-28808/page-2/?recent=498011#post=497787 generator=rsdiscuss&baseurl=https://staging.etf2l.org&feed=forum&forum=off-topic&topic=28808&post=497787 Sun, 25 May 2014 17:48:26 +0200