How to calculate $ \int_0^{\frac{ \pi }{2}} \frac{\sin^{1395}x}{\sin^{1395}x + \cos^{1395}x}\ dx $? Solve the following integrals:  $$ \int_0^{\frac{ \pi }{2}} \frac{\sin^{1395}x}{\sin^{1395}x + \cos^{1395}x}\ dx $$  And it does not follow a specific pattern.
Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$. Evaluate the integral $$\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, \mathrm dx.$$  How can i evaluate this one? Didn't find any clever substitute and integration by parts doesn't lead anywhere (I think).  Any guidelines please?
How can I use a variable in the replacement side of the Perl substitution operator? Few days back while attending a thesis defense, one Professor was asking the defender why does he think, he deserves the degree. I was wondering since then, is there really any general answer for that? or the answer should be given describing my own work and then emphasis them how it is important for research field?  I asked the question here to know your opinion about it and to list down what could be the possible answers to this question. I know SO's policy is &quot;you don't have to register to ask questions,&quot; but there are so many new members and so many hit-and-runs... The majority of the questions I see now are coming from new users (i.e., it's their first question). Although unregistered users should still be able to ask questions, I think we should encourage users to create an account. Here is a couple of ways we could do that:  Explicitly encourage people to get an account: there's currently nothing telling people to do that. Maybe tell people something like &quot;Get an account, it's easy!&quot; and maybe show some of the perks of having an account (and having rep).  Maybe give a few bonus points for registering; and have that enable a basic privilege. It could be taken from one of the , for example.   Of course getting people to register is just half of the battle. I don't know what the stats are on this, but I think there's a number of users who create multiple accounts anyways. We could curb that by:  Having smart(er)? duplicate account detection systems, alerting someone who is about to create a new account and possibly has an account already. Give them a message like &quot;Hey, it looks like you already have  account, are you sure you want to create a new one?&quot;  Create a new privilege (10k? 20k?) that gives users tools to detect possible duplicate accounts, eg: &quot;List users with the same e-mail,&quot; &quot;List users with the same IP addresses,&quot;  etc. (without revealing said emails and IPs). Then the 10k/20k can look at other stuff (e.g.: posts) and refer the new user to a moderator for a possible merge. I'm planning to attend a Californian university as a visiting scholar (not paid by the university) on a J-1 visa, for a total of four weeks; a colleague of mine will do the same, but for 12 weeks. I've been told (personally, ie. no citable sources) that, if you're staying on a J-1, you're effectively a California resident for the time you're there, and, therefore, need a Californian driver's license.  I've read up on this and found  that states:     If you are a visitor in California over 18 and have a valid driver   license from your home state or country, you may drive in this state   without getting a California driver license as long as your home state   license remains valid.      If you become a California resident, you must get a California driver   license within 10 days. Residency is established by voting in a   California election, paying resident tuition, filing for a homeowner's   property tax exemption, or any other privilege or benefit not   ordinarily extended to nonresidents.   I'm not planning to vote in any elections, nor will I own a home there, but I'm not sure what would constitute "privileges or benefits not ordinarily extended to nonresidents".  Do we need to get driver's licenses in California, or can we use our foreign licenses? How about day trips to Arizona or Nevada?  Note: this seems to be related to , but my question is specifically about staying there on a J-1. Even if many interesting similarities between the classical and the quantum mechanical framework have been worked out, e.g. in the subject of deformation quantization, in general, there are some mathematical problems. And in the conventional formulation, you don't want to make things like $\hbar\rightarrow 0$ for the expression $P=-\text i\hbar\tfrac{\partial}{\partial x}$.   In special relativity there are many formulas where one optains the non-relativistic formula by taking the naive limit $c\rightarrow \infty$, e.g.  $$\vec p=\frac{m\vec v}{\sqrt{1-|v|/c}}\ \rightarrow\ \frac{m\vec v}{\sqrt{1-0}}=m\vec v.$$  I wonder if it is know that you can always do that. Is there a formulation of special relativity (maybe it's the standard one already), where the starting assumptions/axioms/representations of objects of discourse involve the constant $c$, and as you take them with you to do all the standard derivations, you always end up with results which reduce to the Newtonian mechanics if you take that limit? etoolbox and tabularx -- problem with `appto` I'm trying to print worksheets. I have already implemented an automatic numbering for subtasks. I use tabularx for my subtasks (see the MWE below).  Yesterday, I tried to define a new macro called \task that has one optional argument which is the amount of points you can obtain for this particular question. I would like these optional arguments to be gathered with etoolbox's \appto command so that I can print them with my \credits command.  So far, this seems to be ind of struggling as tabularx is working on the tables several times. Every time tabularx calculates the width of X columns the points are appended to \credits which leads to an incorrect list of points.  Is there any way I can solve this problem? I'm currently stucked with this issue.  Plus, I don't know why my \task command does work. Why is the counter subtask not affected by tabularx's several trials?  \documentclass{scrartcl}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  \usepackage{etoolbox} \usepackage{booktabs} \usepackage{ltablex} \usepackage{xifthen}  \def\credits{} \newcounter{subtask} \setcounter{subtask}{1} \renewcommand{\thesubtask}{\alph{subtask})} \newcommand{\task}[1][]{%     \ifthenelse{\isempty{#1}}{%         \thesubtask%         \stepcounter{subtask}%     }{%         \thesubtask%         \gappto\credits{(#1)}%         \stepcounter{subtask}%     }% }  \begin{document} \begin{tabularx}{\linewidth}[l]{l@{\hspace*{1ex}}X}     \task[1]    &amp;   TEXT1\\     \task       &amp;   TEXT2\\     \task[3]    &amp;   TEXT3 \end{tabularx} \credits \end{document}       The solution given by David Carlisle below causes a table that is too wide. How can this be undone?  \documentclass{scrartcl}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  \usepackage[a6paper, showframe=true]{geometry}  \usepackage{etoolbox} \usepackage{booktabs} \usepackage{ltablex} \usepackage{xifthen}  \def\credits{} \newcounter{subtask} \setcounter{subtask}{1} \renewcommand{\thesubtask}{\alph{subtask})} \makeatletter \newcommand{\task}[1][]{%     \ifx\hfuzz\@tempdima% %   % tabularx trial do nothing     \else%         \ifthenelse{\isempty{#1}}{%             \thesubtask%             \stepcounter{subtask}%         }{%             \thesubtask%                 \gappto\credits{(#1 BE)\\}%             \stepcounter{subtask}%         }%     \fi } \makeatother   \begin{document}  \begin{tabularx}{\linewidth}[l]{l@{\hspace*{1ex}}X}     \task[1]    &amp;   Some longer text that exceedes the line width.\\     \task[4]    &amp;   TEXT2\\     \task[3]    &amp;   TEXT3 \end{tabularx} \credits  \end{document} How do I hide a process in Task Manager in C#? I have a requirement to hide a process in Task Manager. It is for Intranet scenario. So, everything is legitimate. :)   Please feel free to share any code you have (preferably in C#) or any other techniques or any issues in going with this route.  Update1: Most of the users have admin privileges in order to run some legacy apps. So, one of the suggestion was to hide it in task manager. If there are other approaches to prevent users from killing the process, that would be great.  Update2: Removing the reference to rootkit. Somehow made this post look negative. No homomorphism from $Z_{16}\oplus Z_{2}$ onto $Z_{4}\oplus Z_{4}$. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. IF statement inside a LINQ SELECT Show $\frac{a_{1}+...+a_{n}}{b_{1}+...+b_{n}}$ is between the smallest and largest fraction $\frac{a_{i}}{b_{i}}$, where $b_{i}&gt;0$.  Attempt  Assume the largest is $\frac{a_{n}}{b_{n}}\Rightarrow$  $\frac{a_{n}}{b_{n}}-\frac{a_{1}+...+a_{n}}{b_{1}+...+b_{n}}\Rightarrow $  $\frac{b_{1}+...+b_{n-1}}{{b_{1}+...+b_{n}}}[\frac{a_{n}}{b_{n}}-\frac{a_{1}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}]\Rightarrow $  if $a_{1}&lt;0$, $w=\frac{a_{n}}{b_{n}}-\frac{a_{1}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}&gt;\frac{a_{n}}{b_{n}}-\frac{a_{2}}{b_{1}+...+b_{n-1}}-...-\frac{a_{n-1}}{b_{1}+...+b_{n-1}}$  any hints or solutions? What is the recommended wording for a generic error message Is "laic" an alternative word to the word "layperson"? Closed form for the sequence defined by $a_0=1$ and $a_{n+1} = a_n + a_n^{-1}$ I have 2 equation in terms of $a,b$ and $c$ .  $3a + 10b + 5c =0$ and $4a + 6b + 2c =0$  I need to find a:b:c and answer is $\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}$  I want to know how to get that?  My attempt:  Given equations can be written in form   $$\left ( \begin{matrix}           3 &amp; 10 &amp; 5 \\           4 &amp; 6 &amp;  2 \\           \end{matrix} \right )  \left ( \begin{matrix}           a\\         b\\         c\\         \end{matrix} \right ) =  \left ( \begin{matrix}            0 \\          0\\         \end{matrix} \right ) $$  But here I don't have any idea how to proceed. I can do for 3×3 matrix and I thought it can be done in same way.  Thanks. How to completely remove virtual packages? So, when someone is red-green colorblind, the colors appear the same to them, like this:   And if you're totally colorblind, then things presumably just appear like they would in a black-and-white movie.  However, this isn't how ultraviolet patterns seem to work. Compare how we see this flower to the version where ultraviolet is visible:   The UV pattern is completely invisible here. However, unlike with the red and green, this isn't because yellow and UV are colors that appear identical when you can't see UV. Look at these flowers:     This time the flowers are purple, but the UV pattern is still invisible. Why is that? Shouldn't the UV pattern still be apparent on at least one of the flowers, just in a different color? And on some other flowers, the UV does appear as a different color. So:    Why is the UV invisible only sometimes?  Does it have to do with the flower using iridescent structures to produce color, instead of a pigment? Can this happen with red and green, as well? What is this shell/Bash syntax: someVariable=someValue someCommand Enlarged partition, filesystem still reports old size Singular or plural verb form where subject includes a "parenthetical" element My question arises from , where OP asks whether he should use the singular or plural verb form after "the title, as well as the tone,"  As luck would have it, when I searched Google Books for , I found...        ...and...        At the risk of having this question closed as just peeving, I feel that the first one should have been plural "err", because without the commas it seems to me both "the title" and "the tone" have more or less equal status as "joint" subjects of the verb, making it plural.  On the other hand, I feel the commas in the second example effectively demote "the tone" to "parenthetical" status (the clause is almost incidental). Since this encourages us think of "the title" as the primary subject, it should have been singular "derives".  I'm no great fan of "grammatical rules", but it seems weird to me that the only two relevant examples for this specific wording should both (in opposite ways!) contradict my inclinations.  What's going on? Beamer item balls go missing while compiling in XeLateX since yesterday's TeXlive update, however, the same code works with pdfLaTeX. Any solution, please. Here is my MWE: \documentclass{beamer}  \setbeamertemplate{itemize item}[ball] \setbeamertemplate{enumerate item}[ball]  \begin{document}  \begin{frame} \frametitle{A tested title} text \begin{enumerate}     \item this is an item \end{enumerate} text3 \begin{itemize}     \item this is an item \end{itemize} \end{frame} \end{document}