Wikipedia:Reference desk/Archives/Mathematics/2009 September 1

= September 1 =

Integration
Earlier this  year  our  lecturer  gave  us  an  integral  to  solve,  then  realised  she  had  made  a  mistake,  and  gave  us  an  easier  one.

The Integral  was  ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx,   which  she  changed  to  ʃ { 2 / [ 3X Ln ( 2x)]} dx,  which  equals  ⅓ Ln[ Ln ( 2x ) ] + C

This was  relatively  easy,  but  what  I  would  like  to  know,  is  how  to  solve  ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx,  and  why is  it  that  simply  adding  a  1  to  the  term  in  the  parentheses  makes  it  so  much  harder ? I am  only  a  first  year  Canterbury  ( New  Zealand )  University  student. Thanks. —Preceding unsigned comment added by 202.36.179.66 (talk) 04:08, 1 September 2009 (UTC)
 * Probably because if you use the u substitution method (which I'm guessing was the suggested one) to try to evaluate the second ("mistake") integral, it doesn't work out too well, whereas with $$\int \frac{2}{3x \ln{(2x)}}\,dx$$, if you let u = ln(2x) then you end up getting du = 1/x dx which substitutes in nicely. However if you use the same method with $$\int \frac{2}{3x \ln{(2x+1)}}\,dx$$ and set u = ln(2x+1) then du = 2/2x+1 dx, which unlike the first problem, does not substitute in nicely. Ginogrz (talk) 04:44, 1 September 2009 (UTC)
 * Just to clarify, in the "easier" integral, the solution should be $$\dfrac{2}{3}\left(\ln{(\ln 2x)}\right)$$ (as the u-substitution yields $$\dfrac{2}{3}\int{\dfrac{du}{u}}$$). -- Kinu t /c  04:54, 1 September 2009 (UTC)


 * As far as I can tell, the more difficult integrand doesn't have an indefinite integral in terms of elementary functions.  Dr Dec  ( Talk )    07:42, 1 September 2009 (UTC)


 * How do you determine that? Black Carrot (talk) 22:22, 1 September 2009 (UTC)
 * Risch algorithm has some details, although only in that superficial hand-wavy kind of way that encyclopedia maths articles always seem to gravitate to. Zunaid 09:57, 2 September 2009 (UTC)
 * Also found Differential Galois theory, according to which the problem of solving indefinite integrals using elementary functions is analogous to the problem of finding roots of polynomial equations using radicals. Read the article for more details. Zunaid 10:08, 2 September 2009 (UTC)

Hyperbolic Functions
Given the value of, say, $$\cos \theta$$ one can determine $$\sin \theta$$ and $$\tan \theta$$ by drawing a right angled triangle with the appropriate side lengths. Is there an equivalent for the hyperbolic functions? I know you can use relations such as $$\cosh^2 \theta - \sinh^2 \theta = 1$$ but this isn't what I'm looking for. Thanks. 92.4.122.142 (talk) 17:39, 1 September 2009 (UTC)


 * If you know $$\cos\theta$$ then you only know the angle. You would also need to know the length of one of the sides to determine the triangle unically. If we knew the length of the hypotenuse, say h, then we would know that the adjacent side has length $$h\cos\theta$$ and the opposite side has length $$h\sin\theta.$$ Knowing one of the non-right-angle angles only gives us the triangle up to a scalling. After all: cosine gives the ratio of the length of two sides, and that's invariant under scalling.  Dr Dec  ( Talk )    17:53, 1 September 2009 (UTC)


 * You have picked up on a minor slip up when writing my initial post that has no bearing on either the statement "Given the value of, say, $$\cos \theta$$ one can determine $$\sin \theta$$ and $$\tan \theta$$" or my question. 92.4.122.142 (talk) 18:12, 1 September 2009 (UTC)


 * Ay, ay, ay. No need to be so touchy. I was trying to help! You don't need to use any triangles. If you know $$\cos\theta$$ then you know $$\sin\theta$$ up to sign: $$\sin\theta = \pm\sqrt{1-\cos^2\theta}.$$ If you know $$\cos\theta$$ and you know $$\sin\theta$$ up to sign then you know $$\tan\theta$$ up to sign:
 * $$ \tan\theta = \pm\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} \ . $$
 * If you're trying to fit these around a triangle then you'll want 0 < θ < π/2, and so all the ±'s just become +. The same goes for the hyperbolic trig functions: we know that $$\cosh^2\theta - \sinh^2\theta = 1$$, so once we know either $$\cosh\theta$$ or $$\sinh\theta$$ we can work out the rest with some algebra. Although there won't be a nice circle picture, you'd get branches of a hyperbola. Please, 92.4.122.142, assume good faith; I was honestly trying to help you.   Dr Dec  ( Talk )    18:44, 1 September 2009 (UTC)


 * My apologies. I thought you were deliberately being pernickety about my question just for the sake of it. I see now that was a wrong assumption to make. Thank you for your help and again, I'm sorry. 92.4.122.142 (talk) 18:47, 1 September 2009 (UTC)
 * It does not appear that you have to apologize. It does not appear that your initial post was uncorrect. You were talking of determining sin(θ) and tan(θ), not a triangle. --84.221.68.104 (talk) 22:57, 2 September 2009 (UTC)


 * For trig functions, we can draw the unit circle (x2 + y2 = 1) and then any point on the circle has coordinates (cosθ, sinθ) for some value of θ. The value of θ corresponds to twice the area of the circle swept to our point, which is also equal to the angle from the x-axis.  This let's us draw a right triangle with unit hypotenuse, and one angle θ that has legs cosθ and sinθ.  For the hyperbolic functions we can do a similar thing with the unit hyperbola (x2 - y2 = 1).  Any point on the right branch of the hyperbola has coordinates (coshθ, sinhθ) for some value of θ.  This θ is also twice the area swept out along the hyperbola to that point, but unlike with the circle that doesn't correspond to any angle that I know of, so I think that's pretty much where the analogy ends. Rckrone (talk) 18:17, 1 September 2009 (UTC)
 * Actually I guess you could draw a right triangle with one leg equal to 1, and then the hypotenuse would be coshθ and the other leg would be sinhθ for some θ. θ wouldn't correspond to any obvious property of the triangle though. Rckrone (talk) 18:24, 1 September 2009 (UTC)