I need to prove that: $$ \int^{\frac{\pi}{2}}_{0} \cos^n x \, \mathrm dx = \frac{n-1}{n} \int^{\frac{\pi}{2}}_{0} \cos^{n-2} x \, \mathrm dx$$ for $n \in \{\mathbb{N} &gt; 2\}$ I know that the integrals will not disappear if we take a derivative of both sides of equation even though they are taken on interval with 0. I thought therefore that the only / the easiest way to prove that must be through induction. I have the basis (for n = 3 we get): $$ \int^{\frac{\pi}{2}}_{0} \cos^3 x \, \mathrm dx = \frac{2}{3} \int^{\frac{\pi}{2}}_{0} \cos x\, \mathrm dx$$ (integration of the LHS) $$\left[\sin x-\frac{1}{3}\sin^3 x\right]^{\frac{\pi}{2}}_0 = \frac{2}{3}\left[\cos x \right]^{\frac{\pi}{2}}_0 \to 1 - \frac{1}{3} = \frac{2}{3}$$ And that works. However, I don't know how to do the hard part of induction since I don't even see that type of simple regularity of changes in the integral of $cos(x)$ that would come with changes of $n$. Therefore I'm not sure how to do the proof (which should be easy according to my book).
I am trying to prove  $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$  This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.   $$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$  First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?  Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$  which leads to  $$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$  Since $(n-1)$ is a constant, we can throw it out front of the integral:  $$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$  I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$  $$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$  According to Wikipedia as noted , this simplifies to:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$  Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?  Assuming knowledge of equation 5, I see how to rewrite it as  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x  - (n-1) I_{n} \tag{6}$$  and solve for $I_n$. I had tried exploiting the fact that  $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$  and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$  which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$  but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
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I was wondering if someone experienced a similar issue, and if you could share something about the approach. I attempted a proof of mathematical induction using an arbitrary base case, but was unsuccessful (and hence this question). Below is what I was trying to do and along with my thinking; if anyone can point me in the right direction I'd appreciate it.  The induction I am using: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose $P(0)$. Furthermore suppose that for each natural number $k$, $P(k)$ implies $P(k+1)$. Then $\forall nP(n)$.  What I am trying to prove: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose for $n_0\in\mathbb{N}$, $P(n_0)$. Furthermore suppose that for each natural number $k\geq n_0$, $P(k)$ implies $P(k+1)$. Then $\forall n\geq n_0 P(n)$.  My attempted proof: define $Q(n)$ to be $n\geq n_0 \to P(n)$. Then we wish to prove $\forall nQ(n)$. We induct on $n$.  Base case (for ordinary induction): $Q(0)$ is $0\geq n_0\to P(0)$. Since $n_0\in\mathbb{N}$, $0\geq n_0$ implies that $n_0=0$. Since $P(n_0)$, $P(0)$, which proves the base case.  Inductive step: we want to show $\forall n(Q(n)\to Q(n+1))$. To do this, we assume $k\geq n_0 \to P(k)$ and try to show $k+1\geq n_0 \to P(k+1)$.  Since $k\geq n_0 \to P(k)$, we first prove the case where $k\geq n_0$ and $P(k)$. By the hypothesis of the proof, we see that $P(k+1)$, which proves the case for $k+1$.  This is where I am having trouble: For $k&lt;n_0$, I am unable to show the implication for $k+1$.  So my questions would be: (1) is the overall approach for the proof correct? (2) If so, how might I go on to prove the case when $k&lt;n_0$?  Thanks in advance. (This is not homework, by the way.) I have a cryptography workshop question I'm having trouble with as follows;  Person A creates a cipher $E_k(m)$ which produces a ciphertext from message $m$ using key $k$. The function inside $E$ is kept secret but the length of $E_k(m)$ is known.  Person B recommends "increasing" security of the cipher by instead doing :   $(E_k(m) \oplus m)\ ||\ (E_k(m) \oplus 1111...11)$      Does this in fact increase security of the cipher or increase new problems.  My thinking is, depending on the function within $E$, xoring the output of the cipher with the plaintext message $m$ could expose the key $k$, meaning the extra complexity is for nothing. Am I on the right track, or am I missing something?  I have tried searching for examples of similar schemes and found nothing (which probably means it's not a good scheme) but I need to justify my answer.   Any steers in the right direction would be greatly appreciated, I'm more than happy to do the research myself just unsure what specifically to look for.   Unfortunately the above context is all I have been provided for this question. Date and time panel is greyed out, how do I restore it? Date and time is showing in the login screen (unity-greeter) but not when I login to my session.  When I go to date and time settings the panel is all grey :    Any idea how to recover from this ? Can I safely uninstall games purchased from Steam? Can I uninstall purchased games from my computer and then later download them through Steam without having to re-purchase if I'd like to?   I'm running short on space on my laptop, looking to free up some space and wanted to be sure that I can safely re-download anything at a later time similar to how you can with the Xbox Live marketplace. I have a Surface Pro 4 and an iMac (late 2019). I would like to connect my Surface Pro to the iMac to use the iMac screen as a second monitor.  Is that possible?  What are the requirements to use an iMac as a second screen? How do I ignore ampersands in a SQL script running from SQL Plus? I have a SQL script that creates a package with a comment containing an ampersand (&amp;). When I run the script from SQL Plus, I am prompted to enter a substitute value for the string starting with &amp;. How do I disable this feature so that SQL Plus ignores the ampersand? Viewing all my questions from Android app What is causing disabled options when adding a mesh/curve object to the scene? Sometimes when adding an object to the scene I see that all options (on toolbar or in window after pressing F6) are disabled. Can anyone tell me when can this happen? What is causing it and how to revert it - make it enabled again. It worked for me when I restarted Blender and reloaded the scene. However I would really like to know what can cause it? Is this a correct demonstration for why elements above untriseptium cannot exist? With the confirmation that elements 113, 115, 117, and 118 are indeed fundamental elements that are now to be named on the periodic table, the next question is: what is the highest atomic number possible for an element? Feynman had a go at this years ago, and he derived (according to my limited understanding) that above element-137 (informally denoted 'Feynmanium') the electrons in the nearest orbit about the nucleus would be traveling at a velocity greater than the speed of light. (Note I am considering the Bohr model of the atom in this question.) I wished to demonstrate to my friends why this is and I came up with the following explanation.  According to Bohr's quantum condition, the angular momentum of an electron about the nucleus is given by $$ L=m_evr_n=n\hslash \implies r_n=\frac{n\hslash}{m_ev},$$ where $m_e$ is the mass of the electron, $v$ its velocity, $n$ an integer, and $r_n$ the radius of the $n$th possible orbit. Because we are concerned with the nearest orbit to the nucleus, $n=1$; and thus $$r=\frac{\hslash}{m_ev}.$$   Now, according to Coulomb's law, if the electron orbits about the nucleus the centripetal motion can be described by $$\frac{Ze^2}{4\pi\epsilon_or}=m_ev^2, $$ where $Z$ denotes the number of protons in the nucleus (the atomic number), and $e$ the elementary charge. Solving for $Z$ and substituting in $r$ from above yields $$Z=\frac{4\pi\epsilon_o \hslash v}{e^2}. $$ But what is $v$? Well, the maximum velocity an electron could ever have is the speed of light, and we wish to find the atomic number associated with an orbiting electron traveling at this speed, so we set $v=c$ and obtain our final result of $$Z=\frac{4\pi\epsilon_o\hslash c}{e^2}\approx 137.521,$$ which implies that for $Z&gt;137$, the electrons at a position of $n=1$ in the Bohr model would have a velocity $&gt;c$; and thus the highest atomic number achievable on the periodic table is 137.  Again, I just want to make sure this is a correct method for deriving element-137 before presenting. Perhaps one could explain how relativity plays a role here. I know Feynman used the Dirac equation to get this result...so could anyone (subsequently of course) expatiate on this in a simplistic manner? Thanks! How is momentum conserved in this example? EDIT: Summary  I think I've found the best definition of what I want thanks to the discussion with @M. Tibbits:  When I open or refresh the SO, I get a notification of new activities that are of interest to me. Today these are (hard-coded): answers to my questions and comments addressed to me. I want a bit more control over it, that's all.     Yes, I've read  and , but I think I add new (good :) ) reasons for re-asking it:  The first question was answered before this so handy "inbox" notification feature was implemented. So now using favorites for "watching" a question is wrong, IMHO, for 2 reasons:   it splits essentially the same functionality (of getting notified when there is an activity you are interested in) between 2 very different user interfaces it's cumbersome - I need to: 1) add question to favorites; 2) click on my profile dropdown; 3) click on the "favorites" line; 4) see all my favorites with recent activity; 5) select the one I watch; 6) repeat this process occasionally to get updates; and 7) possibly remove the question from favorites at the end. All this as opposed to this little red button when there are news.   Favorites are different in nature that watching. I add a question to Favorites when I find the answers useful to me in my work, so I may want to have it handy in the future. By watching an interesting question I will be able to decide whether I want it in faviorites, after I see the answers.  EDIT: I see that I was not clear enough.  I want to add a question to Favorites when:   it has some good answers the question and answers are relevant in my work I want to keep them handy as my long-term (persistent) knowledge base   I want to watch a question for short while (matter of days/weeks) when:   it is interesting / relevant I want to follow it to see what are the answers. Then I can decide whether I want it in my favorites   As you see, the main differentiators are good answers and long/short term. For short-term, automatic notification is better.  Bottom line, all I ask is "watch this question" checkbox (which may expire after say a week if there is a need to free resources) I am trying get an unused server up and running, it booted fine but i was greeted with the login window and none of the known passwords seem to work. The server I am trying to get to work is not a Domain Controller and has not been used in a long time.  OS: Windows 2003 Server R2 Enterprise Edition  I tried Ophcrack XP (there is no WinServer version of Ophcrack) - it couldn't locate the partitions with hashes. I tried mounting the partitions myself (#mount /dev/hda*x* /mnt/ntfs), but it gave me an error saying /dev/hda*x* didn't exist.  Next I tried booting into a Ubuntu 10.04.3 LiveCD so that I could install and use chntpw - no go, Ubuntu Live wouldn't boot.  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