Evaluate  $\int_{\frac{25\pi}{4}}^{\frac{53\pi}{4}}\frac{1}{\left(1+2^{\sin x}\right) \left(1+2^{\cos x}\right)}dx$  My Attempt:  I tried using the property that $\int_{a}^{b}f\left(a+b-x\right)dx=\int_{a}^{b}f(x)dx$ and properties of periodic functions but ended up with a mess. The answer is $\frac{7\pi}{4}$
How to evaluate the integral $$\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?$$
Is there such a thing as charge density in quantum mechanics? Is there such a thing as a time-varying charge density in quantum mechanics? I am looking at the tungsten filament in a light bulb. Am I allowed to say that there is a wave function which describes that filament, and that in some sense there is a time-varying charge density (which some people call a probability) associated with the square of that wave function? There are electrons which have a probability distribution, and there are nuclei which are vibrating within the lattice, but all in all is there, or is there not, a time-varying charge distribution associated with that glowing filament? And if not, why not?  EDIT: There seems to be some question as to just what kind of charge-density fluctuations I am talking about. Let me clarify: in a hydrogen atom in the ground state, I would say there are NO charge density fluctuations. But if there is a superposition of the ground state and a 2p state, I would say the charge distribution fluctuates sinusoidally at the difference frequency of the two states. THAT is the kind of charge density fluctuation I am asking about for in the tungsten filament. Does it or doesnt' it?  I hope this clarifies the intent of my question.   RE-EDIT: In a separate (and rather hard-fought) discussion, it appears that people who know more than me agree that there is indeed an oscillating charge distribution in the hydrogen superposition as described above (see  ).  So if there is an oscillating charge in the hydrogen atom, why wouldn't there be oscillating charge in a heated tungsten filament? What canon evidence is there that Stormtroopers are/are not still clones? Reading this article about , my nerd-rage started flaring as they claimed that Stormtroopers were still Fett-Clones 20 years after the rise of the Empire.  As I started thinking about it, I realized that everything I had read saying that Modern-Day Stormtroopers not being clones was from the EU - C Level books or games.   One of the quotes often cited for backing the idea that the Stormtroopers are clones comes from Princess Leia:     Aren't you a little short for a Stormtrooper?   But through the movie, paying close attention, Stormtroopers are often shown to be slightly different heights. One even hits his head in the doorway while others walk through uninjured.                                  Is there any G Canon fully canon (non-Legends/Infinities) evidence to support the idea that Stormtroopers are, or are not still clones at the time of Episode IV? How to create a bootable USB stick? I am looking at this:    I need to create a boot disk to test Ubuntu to make sure it will run on a PC (Compaq Mini CQ10-120LA) I was given. I can create the boot disk off of a Mac (in English) or Windows (but Windows is in Spanish and foreign to me).   Questions:     What format do I choose for the USB stick? (I wish the instructions stated this). What is Dash? (Will I know when I run the installer?) Can I do this from a Mac or Windows computer? Or only from Ubuntu? No route matches "/users/sign_out" devise rails 3 how to prove that $\ln(1+x)< x$ How do I go about evaluating the following by contour integration?  $$ \int^1_0 \frac{dx}{(x^{2} - x^{3})^{1/3}} $$  The question does not fit in the standard form of : $\int^{2\pi}_0$ or $\int^\infty_{-\infty}$  etc.  I tried substituting $ x = cos(\theta) $ but got stuck. If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum. If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \dfrac{n-1}{n+1}$, find $a_n$ and the sum.  By definition,  the sum of the series is the $\lim n\to\infty$ of it's $n^{th}$ partial sum.  $$\text{ Sum = } \lim_{n\to\infty} = \dfrac{n-1}{n+1} = 1$$  I am asked to find $a_n$. How do I do this? What is the procedure? Setting proxy for apt from terminal Possible Duplicate:          I have tried changing my proxy settings in a terminal as:  export HTTP_PROXY=http://10.1.3.1:8080   and  export http_proxy=http://10.1.3.1:8080   but when I try to install a new package or update apt-get, apt-get starts displaying messages from which it seems it is trying to connect to a previously set proxy:  sudo apt-get update   0% [Connecting to 10.1.2.2 (10.1.2.2)] [Connecting to 10.1.2.2 (10.1.2.2)   I have tried setting the proxy via bashrc file but that din work either. As far as I remember 10.1.2.2 was set using GNOME GUI but I don't have access to the GUI right now so I am trying to set it from terminal. Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$ Prove that $[0,1]$ isn't homeomorphic to $\mathbb{R}$  My first thought is that there can not be a continuous bijection, $f$, from $[0,1]$ to $\mathbb{R}$ because a continuous function that maps $[0,1]\rightarrow \mathbb{R}$ must be bounded so there can't be a surjection. So that would then be the proof.   Though I believe I am incorrect in my thinking because a hint for the assignment says to use the intermediate value theorem. That is "Suppose $f:[a,b]\rightarrow\mathbb{R}$ is continuous. If $f(a)&lt;\delta&lt;f(b)$ or $f(b)&lt;\delta&lt;f(a)$ then $\delta=f(c)$ for some $c\in[a,b]$".   Why is my first thought wrong and how is the IVT useful? How to understand Cauchy's proof of AM-GM inequality(the last step) The AM-GM inequality: $$a_1a_2\cdots a_n\leq\left(\frac{a_1+\cdots + a_n}{n}\right)^n$$  the trivial case: $a_1a_2 \leq \left(\frac{a_1+a_2}{2}\right)^2 $ is self-evident.  then cauchy use this fact repeatedly. He get: $$a_1a_2\cdots a_{2^m} \leq \left(\frac{a_1+a_2+\cdots +a_{2^m}}{2^m}\right)^{2^m}.\tag{$\ast$}$$  This is easy to understand, and natural, but next step:  he let  $$b_1 = a_1,\ b_2=a_2,\ \ldots,\  b_n=a_n,\  b_{n+1}=\cdots=b_{2^m}=\frac{a_1+a_2+\cdots a_n}{n}$$  replace $a_1, a_2,\ldots ,a_{2^m}$ with $b_1, b_2, \ldots b_{2^m}$ in $(\ast)$ and simplified both left and right. He get the answer. I can repeat his proof, but I don't know why Cauchy can got this idea. How does he got it? In which ways did scanning Nero's ship enhance/change Starfleet technology? I've heard the explanation that the Enterprise from 2009's Star Trek looked so much more advanced was because data gathered from the Kelvin initially scanning Nero's ship when it first appeared enabled them to upgrade their tech. I'm wondering if they've ever detailed exactly what technologies were advanced or enhanced or discovered. Can I install Android on my non-Android device? Problem:   Prove that a set of nine consecutive integers cannot be partitioned into two subsets with same product   My attempt:  This problem would be solved if it could be proven that 9 consecutive numbers cannot exist such that all the numbers are of the form $2^a  3^b  5^c  7^d$ where $a,b,c,d$ are integers.  A suitable hint would be appreciated, I don not want a complete solution.  Edit:  If there existed a prime factor greater than $7$ of any number present in the set, then its multiple cannot exist b/c $22-11&gt;9$  Edit 2: As mentioned by Erick, my reasoning is not valid for numbers $1-9$ and $2-9$, let this be an exception. How do you get a bad transcript past Ph.D. admissions? Tiny navigation buttons and tags on responsive mobile I know it will be hard for you to tell on desktop because the resolution is so high, but the clickable text in the upper left, the icons on the upper right, and the navigation buttons at the bottom are all tiny. I have to squint to read the tags, I can't reasonably tap any navigation button except for Next without zooming, and selecting the achievements dialog, etc. is all very difficult. It seems to me that this would not be terribly difficult to change, so I am posting this feature request. I'm using a relatively large Samsung phone (Note 8), so I can imagine it would be worse on smaller phones.  Edit: I see from posted images that the responsive design on other phones looks nothing like mine. I'm wondering if it has something to do with the high resolution  where it thinks it is bigger than it is, or it's picking the wrong sizes because it's a tiny bit over some size threshold, in which case this would be a bug. What is the "Install this third-party software" box in the installer and what third-party software is installed? Add parenthesis as an accepted Markdown ordered list delimiter The Markdown editor knows that an empty line, followed by a number, followed by a period, followed by a space, is a list item.  This works great It's almost magic  The problem is that I almost always use a close parenthesis for my lists. I think it's a pretty common practice, but the problem is that the editor / Markdown parser doesn't accept parenthesis as an acceptable list delimiter:  This fails horribly This is on the next line I swear What does "period" mean when someone says "Sth... Period"? I often hear US citizens say something like this:     &lt;Some phrases>. Period.   What does period mean when it appears in a sentence by itself in this type of situation? Equivalent Resistance? How do you find the equivalent resistance here? (I realized it's a Wheatstone bridge, but it's unbalanced, so that doesn't help.) I tried using Kirchhoff's laws, but there is no cell given. How can I download an APK file from the Play Store?