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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\begin{equation*} xJ''+J'+xJ=0 \end{equation*}\] \[\begin{align*} J(x)&=\frac2\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ J'(x)&=\frac2\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)+\,\mathrm du\\ \\ J''(x)&=\frac{2}\pi\int\limits_0^{\pi/2}\frac{\partial}{\partial x}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du\\ \\ \end{align*} \]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[ \begin{align*} &xJ''+J'+xJ\\ &=\frac{2x}\pi\int\limits_0^{\pi/2}\sin^2(u)\cos\left(x\sin (u)\right)\,\mathrm du+\frac{2}\pi\int\limits_0^{\pi/2}\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du+\frac{2x}\pi\int\limits_0^{\pi/2}\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x\sin^2(u)\cos\left(x\sin (u)\right)\sin(u)\sin\left(x\sin (u)\right)+x\cos\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x(1\sin^2(u))\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ &=\frac{2}\pi\int\limits_0^{\pi/2}x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\,\mathrm du\\ \end{align*} \]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i think i need to find [?]\[\begin{equation*}\frac{\partial }{\partial u}\Big[\quad?\quad\Big]=x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]
 one year ago

malevolence19 Group TitleBest ResponseYou've already chosen the best response.0
I wish I was in this class :/
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
am i thinking about this problem the right way?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
hmm wolfram tells me \[\begin{equation*}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]=x\cos^2(u)\cos(x\sin(u))\sin(u)\sin\left(x\sin (u)\right)\end{equation*}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
I am meant to be able to work that bit out in my head?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
its just the product rule in reverse but is kind hard to make that step
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
The work appears correct to me here. I can't say I'd be able to get that antiderivative easily though, in my head... lol
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
the wolfram stepbystep solution starts by assuming we can tell the factors in the product rule, after we have worked out these , the problem is simple, but i have trouble seeing the factors
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\begin{align*}&=\frac{2}\pi\int\limits_0^{\pi/2}\frac{\partial }{\partial u}\Big[\cos(u)\sin\left(x\sin(u)\right)\Big]\,\mathrm du\\ &=\frac{2}\pi\cos(u)\sin\left(x\sin(u)\right)\Big_0^{\pi/2}\,\mathrm du\\ &=\frac{2}\pi\left[\cos(\tfrac\pi2)\sin\left(x\sin(\tfrac\pi2)\right)\cos(0)\sin\left(x\sin(0)\right)\right]\,\mathrm du\\ \\ &=0\\\end{align*}\]
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
\[ \begin{align} \frac{\partial}{\partial u} \left( \cos u \; \sin ( x \sin u ) \right) &= \frac{\partial}{\partial u} \left( \cos u \right) \sin (x \sin u) + \cos u \frac{\partial}{\partial u} \left( \sin ( x \sin u ) \right) \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin (x \sin u) } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ x \cos u \cos (x \sin u) } \\ \text{Let } k = x \sin u; \quad k' = x \cos u. \\ &= \color{#aa0000}{ \neg \sin u } \color{#00aa00}{ \sin k } + \color{#00aa00}{ \cos u } \; \color{#aa0000}{ k' \cos k } \end{align} \] I think it looks a little clearer when you replace the weird composition of functins and find the derivative of the inner function / replace that...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
that is a good method letting k= x sin u , because it is common to both terms thanks. P.S. \(\neg\) is not the same as\(\)
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.2
ooh, I see what that is now. I kept thinking \neg was a negative sign and the bent end was just a weird latex thing. i remember seeing it in something else now.. lol :P You're welcome, and thanks!
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
negation is used in logic and sets, for example if the universal set is U={1,2,3,4,5,6,7,8} and a subset is S={1,2} the negation of S ¬S={3,4,5,6,7,8}
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.0
In settheory it's called a *complement* rather than a negation, which is typically used in logic.
 one year ago
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