One way that I tried to solve an integral was to divide integrand by $\cos^3x$ para para obter  $$ \int_{0}^{\pi/2}\dfrac{\sec^3x}{1 + \tan^3x}dx $$ Through software, I saw that the result involves $\tanh^{-1}\bigl(\frac{1}{\sqrt{2}}\bigr)$, but I do not know how to get there.
How do I integrate the following $$\int \frac{dx}{\sin^3 x + \cos^3 x}$$ ?  It appears that I am supposed to break this up into $(\sin x + \cos x)(1-\cos x \sin x)$, but the next thing to do is not apparent to me.
Bound for the correlation of three random variables Can you use Magical Secrets to choose high level spells when multiclassing as a Bard? Say you have a Wizard 7/Bard 9. You gain your 17th level and choose to gain the level in Bard (which grants you Magical Secrets). According to the Multiclass Spellcaster: Spell Slots per Spell Level table, you now have a single 9th level spell slot. Can the spells you choose from Magical Secrets be 9th level spells even though you can only cast up to 5th level Bard spells? What is the best prepaid SIM card for data in France? I'm going to travel to Paris in two weeks for 5 days. I'd like to buy a prepaid SIM card for my phone, no calls or SMS needed - only data is sufficient. I've looked through French mobile operators and found nothing really acceptable. For example, in Finland I bought SIM card for €5 with 3GB of data.   So, the question is - what is the best SIM card for me, assuming that 1GB is enough (considering the prices) and that I am actually will be able to use it (I've read that some operators turn on mobile data only after 5 days which is unacceptable for me considering that I'll be there for 5 days only)? If we have 2 batteries one of emf x and the other is of emf y and we connect them in series we get an effective emf of x+y.  But what if we connect them in parallel, how to calculate the emf now? How to make UV loft like 3ds Max in Blender using curves? I am try to make a car Body using the technique like UV loft in 3ds Max.  How to apply this technique in  Blender using curves or closer to this technique? I have seen this technique in a Blender tutorial, but I forgot about it. Can I uninstall and reinstall Ubuntu Software Center? Make the enter key send message on iPad Proof that no polynomial with integer coefficients can only produce primes Doing a discrete math review and am trying to solve problem 1.6 in the text found here:  - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here:  For n = 40, the value of polynomial $p(n)::=n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers.  In particular, let $q(n)$ be a polynomial with integer coefﬁcients, and let $c::=q(0)$ be the constant term of q.  (a) Verify that $q(cm)$ is a multiple of c for all $m \in Z$  Proof: Since q is a polynomial, it will be of the form: $q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c$  If $x = cm$, then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all $m \in Z$.  b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, $N$, there are infinitely many $q(n) \in Z$ that are not primes.  Proof: Continuing from the result found in (a), it is easy to see that for all $m \in Z$, there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime.  c) Conclude that for every nonconstant polynomial, q, there must be an $n \in N$ such that $q(n)$ is not prime. Hint: Only one easy case remains.  c = 0 is trivially easy to prove using the (b) above.  I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime. + or @ mark after running 'ls -al' In Mac OS X, run 'ls -al' gives me something like this.   drwxrwxrwx+  4 smcho  staff     136 May  5 09:18 Public drwxr-xr-x+  6 smcho  staff     204 Feb  1  2010 Sites drwxrwxrwx   9 smcho  staff     306 Feb  2  2010 backup drwxr-xr-x@ 36 smcho  staff    1224 Sep  4 22:51 bin    What's the + or @ at the end of the first column means? Is this unique to Mac, or common in UNIX?   ADDED  After Michael Mrozek's answer, I ran 'ls -ale' to get the following.   drwx------+ 66 smcho  staff    2244 Aug 30 13:40 Library  0: group:com.apple.sharepoint.group.3 allow search  1: group:everyone deny delete drwxr-xr-x   3 smcho  staff     102 Sep  4 15:01 Mail drwx------+ 13 smcho  staff     442 Aug 28 17:55 Movies  0: group:everyone deny delete drwx------+  6 smcho  staff     204 Jul  9 09:37 Music  0: group:everyone deny delete drwx------+ 11 smcho  staff     374 Aug 28 16:55 Pictures  0: group:everyone deny delete drwxr-xr-x   3 smcho  staff     102 Mar 18 15:43 Projects drwxrwxrwx+  4 smcho  staff     136 May  5 09:18 Public  0: group:everyone deny delete drwxr-xr-x+  6 smcho  staff     204 Feb  1  2010 Sites  0: group:everyone deny delete   What those appended messages mean? Why do I have them for some of the files? I don't remember doing anything particular for them. "At the time" versus "at that time" Possible Duplicate:          There are a lot of really screwed up configurations out there, because people didn't take a minute to sit down and think about something as simple as the name for their Active Directory forest. I'm asking this question to self-answer so that this knowledge can be more readily available. If you think you have a better answer, by all means add your input (but be prepared to defend it!)   So, what should I name my new Active Directory? How to Mount Guest DOS OS with HOST OS.? I am newbie to VM concept. actually i have installed DOS6.22 under virtualbox and using 32bit windows 7 OS as HOST Operating System, now i want to copy some folders/files from HOST OS to DOS6.22 Guest OS. how can i copy.? Which way is up? (electric outlet) Password protect folder without disk images? Is this possible? I know you can protect a .dmg file, but is there a more convenient solution, by using a third party application? Everytime I try to sudo something at the terminal, I get a warning message:  alexandre@XPS-15Z $ sudo mount file.iso /mnt/ -o loop [sudo] password for alexandre:  no talloc stackframe at ../source3/param/loadparm.c:4864, leaking memory   It's just annoying to get that all the time. How do I solve that? I'm on Ubuntu 14.04 (upgraded from 13.10) How do I create a line graph which ignores zero values? I want to create a graph showing trends over time, calculated from a formula of various data.  However, some data is missing, resulting in zeroes in the calculated data which the chart uses.  When plotting this data against as a line graph, it's drawing a line down to the axis, which I don't want.  I'd like either a gap in the line, or skip the data point and draw from directly previous to next.  i.e. given this data:  00.0    61.6    39.5    74.6    86.8    72.0    66.5    99.5    85.6    80.6    95.0    122.4 81.0    83.1    88.2    00.0    94.3    60.5    00.0    99.5    95.8    86.2    00.0    111.4 84.1    110.8   78.9    62.0    75.9    102.2   89.9    87.1    81.6    73.5    00.0    116.9   All the 00.0 items mean "no data", so I'd like them ignored (not treated as 0 value).   This is the chart I currently have:   This is how I'd like the chart to come out:   Again, I don't mind if the dotted lines are blank or solid, so long as they don't go down to zero.  The data is dynamic/calculated, so I can't just say =AVERAGE(prev,next) since I don't know where missing data will occur in advance - I need a solution which takes this into account.  Any ideas? Does "std::size_t" make sense in C++? In some code I've inherited, I see frequent use of size_t with the std namespace qualifier.  For example:  std::size_t n = sizeof( long );   It compiles and runs fine, of course.  But it seems like bad practice to me (perhaps carried over from C?).  Isn't it true that size_t is built into C++ and therefore in the global namespace?  Is a header file include needed to use size_t in C++?  Another way to ask this question is, would the following program (with no includes) be expected to compile on all C++ compilers?  size_t foo() {     return sizeof( long ); } FrameLabel and OverTilde One of my associates had their Hotmail account broadcasting spam, and presumably hacked. They managed to change the password, but after that, the account got locked; Hotmail asks you for a phone number, and they're supposed to call you with a code.  This person requested the code several times, with different numbers (even two of mine!), but never received any call with the confirmation code.  I referred them to the full-length process, which includes answering the security question; apparently, they forgot their question (having entered it over a decade ago and never reviewed it since).  What options, if any, exist to restore access to this person's account? They only want to grab their contacts; they plan to operate full-force on GMail going forward. Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$