| andrew123 | can anyone integrate using substitution: sin[(sqrt)x] / (sqrt)x |
| Safrole | u = sqrt(x), so du = 1/(2sqrt(x)) dx |
| Copter2 | looks like a sin there though :o |
| Safrole | thus we get int( sin(sqrt(x))/sqrt(x) dx ) = 2*int( sin(u) du ) = -2cos(u) = -2cos(sqrt(x)) |
| TheBlueWizard | equivalently, x = u^2, dx = 2 u du, and then substituting is pretty straightforward |
| [n01d] | hi guys, whats the function/algorithm to calcalculate sine? |
| andrew123 | safrole: if u = sqrt(x) wudnt we get sinu/u du |
| Safrole | if u = sqrt(x) then du = 1/(2sqrt(x)) du = 1/(2sqrt(x)) dx, sorry so then 2 du = 1/sqrt(x) dx so then we get int( sin(sqrt(x))/sqrt(x) dx ) = int( sin(u) du ) = 2*int( sin(u) du ) sorry I forgot the 2 |
| andrew123 | but we are integrating in regards to u.. so dont we need du/dx = 1/2sqrt(x) then dx = 2sqrt(x).du |
| Safrole | andrew you're mixing things here that need nod be mixed we want a substitution for 1/sqrt(x) dx I found one |
| bsmntbombdood | is >, < defined on C? |
| Safrole | < . > you mean? An inner product? |
| [n01d] | can someone help me whats the manual to solve sine. not using calculator |
| bsmntbombdood | no, greater/less than |
| Safrole | C is *not* an ordered field |
| Eclipsor | [n01d]: ? |
| Safrole | you can impose a so called lexicographical order |
| Eclipsor | solve sine in what way |
| Safrole | but that still has some nuances But in general, no > and < are not defined for C |
| shinygerbil | [n01d]: do you mean like the Taylor series of sin(x)? |
| bsmntbombdood | ok |
| Safrole | the Taylor series would require knowledge of cosine |
| shinygerbil | hehe, true |
| Safrole | that's kind of circular. |
| bsmntbombdood | Safrole: no it wouldn't |
| shinygerbil | no, wait well, kinda |
| bsmntbombdood | cos(0) is easy to find |
| shinygerbil | actually doing a Taylor expansion on sin(x) would |