I have to calculate this Integral -  $$\int _0^{\frac{\pi }{2}}\:\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}dx$$  I have no idea how to start, Hint someone ? Thanks.
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?
I've recently started playing Minecraft, and I only have a little experience with enemies (I've been playing with the peaceful setting so I can figure out what the heck I'm doing).  When I did play some survival mode, I happened to be at the top of a pretty large cliff when night hit, and this seemed to protect me - the baddies just piled up at the bottom of the cliff.  However, it seems like this probably only worked due to pathing flukes.  The terrain behind me was pretty easy going, so I suspect if enemies spawned in the right areas, they could still come up and visit.  That said, how can I set up a quick protective shelter for the first night or two?  The days don't seem to last very long, so I don't expect I would be able to generate a lot of high-end materials, tools, etc. to work with.  Also, the zombies caught fire nicely when the sun came up, but there were still some spiders and some other things (creepers? - they were explosive) that didn't go away at daybreak.  Do I need multiple exits for my shelter or some other way to avoid running into all the enemies outside the front door that have accumulated overnight? What is the best way to restrict access to adult content? I bought my kids a PC and installed 12.04 (Unity) on it.  The bottom line is, I want my children to use the computer unsupervised while I have confidence they cannot access anything inappropriate.  What I have looked at:  I was looking at Scrubit a tool which allows me configure my wifi router to block content and this solution would also protect my other PC and mobile devices.  This may be overkill as I just want the solution to work on one PC.    I also did some Google searches and came across the application called .  My experience of OSS is that the best solutions frequently never appear first on a Google search list and in this case I need to trust the methods therefore my question is very specific.   I want to leverage your knowledge and experience to understand What is the best way to restrict adult content on 12.04 LTS as this is important to me.  It maybe a combination of things so please don't answer this question "try this or that", then give me some PPA unless you can share your experience of how good it is and of course if there are any constraints. Develop rich API exposing Stack Overflow functionality I would love to see an actual native iPhone app for Stack Overflow, Server Fault, Super User, etc.  I've actually developed a number of iPhone apps myself, and would certainly be interested and willing in doing the bulk of the legwork, but it would be necessary for the SO team to expose a real API, instead of simply RSS feeds.  Maybe if there's enough demand behind this, we can convince the SO team to build a rich API, which could be useful for many other applications, not just an iPhone app. $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$ Let $p$ be an odd prime, and let $g$ be a primitive root modulo $p$. Prove that $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$.  I've seen this solution on the math stack exchange but most of the solutions use quadratic reciprocity. Unfortunately, I skipped that section because it seemed very hard, but the proof that  $g^\frac{p-1}{2} \equiv -1 \ (mod \ p)$ seems easy with it.  Is there another way to solve this This question is proposed in the chapter of Finite Abelian Groups.  DEF:In elementary number theory, an integer $g$ is called a primitive root for the modulus $n$ if $\mathbb{Z}_n^\times$ is a cyclic group and $[g]_n$, is a generator for $\mathbb{Z}_n^\times$. Should smilies be turned into images in questions and answers? A lot of mention has been made of the seriousness of Stack Overflow (and presumably Server Fault). Would it be a good idea to turn  :-) into actual images as per  in questions and answers? How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$ Question: Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}&gt;0$, and for all $n \in \mathbb{N}$ we have:  $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$  Prove that :  $\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.  My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x&gt;0$; then we have $$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$ But I can't see how to prove the limit of $\{a_{n}\}$ exists. Thank you for you help! How to say that the phone is held between shoulder and ear? What are common phrases an English native speaker would use for the pose of speaking on the phone where you use your shoulder to keep the receiver against your ear (and mouth), so your hands are free?  I have some ideas, but they all sound wooden, and I don't think I've ever heard anyone describe this pose in words. It's also pretty much impossible to google for a pose or look it up on leo.org or other dictionaries.  OK, since people keep asking for what phrasings I found, I'll provide two, though I was hoping I could avoid influencing people who answer here by not providing them: "balance on one's shoulder", "held/nestled/sandwiched between shoulder and ear". Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right. Two icons on Unity Launcher on non-standard application launch I have an application which doesn't come from any Ubuntu repo (neither official, nor PPA) but available as tgz and supposed to be unpacked and ready to use. To be precise it is .   The problem is the  application when launched shows its own icon in Unity Launcher. After making a custom .desktop file and placing shortcut onto Unity Launcher (by dragging from the dash) and launching it I see two icons - one placed by me and another one with real application.  This is the .desktop contents:  [Desktop Entry] Version=1.0 Type=Application Terminal=false Exec=/opt/LightTable/LightTable Name=Light Table Icon=/opt/LightTable/core/img/lticon.png   One more thing. The /opt/LightTable/LightTable is bash script file and eventually /opt/LightTable/ltbin is being launched. I think this is the problem but have no idea how to fix the problem. I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $U(t)$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $E$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar} $$ where $\lvert E, \alpha \rangle $ are the orthonormal eigenkets for the $E$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$ U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$  I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $\lvert E\rangle = \lvert p\rangle$ and $p = \pm\sqrt{2mE}$, he chooses the propagator as a function of the momentum eigenvalues $p$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $-p$, and the bounds switch from $0$ to $\infty$ such that: $$U(t) =  \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $\lvert p\rangle = \lvert E, + \rangle$,  $\lvert -p\rangle = \lvert E, - \rangle$, $E = p^2/2m$, and $dp = \pm mdE/\sqrt{2mE}$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}. $$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $ m/\sqrt{2mE} $ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues? Let $a\ge 2$ be an integer. Show that for positive integers $m,n$, we have $a^n - 1$ divides $a^m - 1$ if and only if $n$ divides $m$.   I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions.  This is what I have so far: Let $m = nq + r$, then $a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1$. Since $0\le r\lt n$, $r = 0$ then $a^m - 1$ = $a^{nq} - 1$ so $a^m$ = $a^{nq}$ . Thus $m = nq$ hence $n\mid m$. People are born and grow up and die of old age in a matter of 5 days. The elders (5 days old) train children and send the most promising one on a trek across the dessert to a monolith that is the cause of the rapid aging to hopefully disable it.   I saw it on a reel to reel in the mid 90's. It looks like it was made in the late 70's or 80's. The cast was small. A few men played the elders seated in a row of chairs and the young boy who had to set out to save them who matures into a man and finally an elderly man.  I've been trying to figure out the title for years. I was wondering if someone could critique my argument here. The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday.  My argument is that there are 23 choose 2 ways times $\displaystyle \frac{1}{365^{2}}$ for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday. This is just 365 permute 21 times $\displaystyle \frac{1}{365^{21}}$. To summarize:  $$\binom{23}{2} \frac{1}{365^2} \frac{1}{365^{21}} P\binom{365}{21}$$ Let's say I have a (comparatively) lovely-looking document in LaTeX, full of lovingly typeset, (relatively) complex equations. Now, let's say some barbarians come along one day and decide that the document has to be put into Microsoft Word (2007). ...after the usual mourning period associated with such events, let's say I value my job (more specifically, the bread it provides) enough to get all the text and tables formatted and references organised into the Word document. –...related questions  and – Now I'm looking at the equations with fear and dread. One option of course is to just lift screenshots from the original document, but this is painstaking if I need to refer to parts of the equation in the text. Also, I might need to edit equations on the fly. Anyone know of a free application which allows embedding LaTeX math into MS Word? I've looked at  and  which do roughly what I want... they build LaTeX images from source and embed them into the Word document, allowing to edit the source later... but both are commerical. ...any help in these troubling times will be greatly appreciated. EDIT: Just a note that  offers a 30-day free trial and is working out really nicely... but still, it's not free. Might be a good solution for those with short-term needs, or money. I was surprised to discover my dictionary had this entry for dilemma:     a situation in which a difficult choice has to be made between two or more alternatives, esp. equally undesirable ones   The notion of dilemma meaning two or more flies against what I was taught about the word. The very idea of a  is specifically based on the number two.  Has my dictionary merely updated its definition to encapsulate the many people who use dilemma for more than two equal choices? Or was someone in my youth being unnecessarily pedantic? I'm looking for some sample code that will sort the list items in an HTML list by alphabetical order.  Can anyone help?  Here is a sample list for people to work with:  &lt;ul class="alphaList"&gt;     &lt;li&gt;apples&lt;/li&gt;     &lt;li&gt;cats&lt;/li&gt;     &lt;li&gt;bears&lt;/li&gt; &lt;/ul&gt; Integrating trigonometric function problem $\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx$ \begin{eqnarray*} \int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &amp;=&amp;  \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\ &amp;=&amp; \int \frac{3\tan x +2}{2\tan x +3} dx\\ &amp;&amp; u = \tan x \text{ and } du = \sec^2 x \ dx  \end{eqnarray*} Am I going in the right direction with this one? It seems like not. Is it possible to create an  shader in Cycles? Specifically for something like , , , etc.. reports embedded in visualforce not showing I have a few reports embedded in visualforce page using iFrame. Due to this new release of Spring '14, iFrame has stopped working and now these visualforce pages show blank screens. Can anyone help me with a work around? inequality with three variables and condition