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).
42547
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   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
119803
I'm trying to prove that $\frac{\mathrm{d} }{\mathrm{d} x}\ln x = \frac{1}{x}$.  Here's what I've got so far: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &amp;= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align} $$ To simplify the logarithm: $$ \lim_{h\to0}\left (1 + \frac{h}{x}\right )^{\frac{1}{h}} = e^{\frac{1}{x}} $$ ^This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{1}{xh}}$, but I can't work out how to get from the above line to that. $$ \lim_{h\to0}\left ( \left (1 + \frac{h}{x}\right )^{\frac{1}{h}}\right )^{h} = e^{\frac{h}{x}} $$ Going back to the derivative: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &amp;= \lim_{h\to0} \frac{\ln(e^{\frac{h}{x}})}{h} \\ &amp;= \lim_{h\to0} \frac{\frac{h}{x}\ln(e)}{h} \\ &amp;= \lim_{h\to0} \frac{h}{x} \div h\\ &amp;= \frac{1}{x} \\ \end{align} $$  This proof seems fine, apart from the middle step to get $e^{\frac{1}{x}}$. How could I prove that part?
68284
closed form of $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}$$  for $a,b&gt;0$  n=1 we get  $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}$$  n=2 we get  $$\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}$$  but what the integral for n ????  I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all
19917
Why do we have, for every $n\in\mathbb N$, $$\int_0^\infty\left(\prod_{k=1}^n \cos\left(\frac{x}{k}\right)\right)\frac{\sin{4x}}{x}dx\approx\frac{\pi}{2}\ ?$$
4558
Show that $$\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right).$$   I don't have any idea how to start, but .
47462
Assume that $x+\frac{1}{x} \in \mathbb{N}$. Prove by induction that $$x^2+\frac1{x^2}, x^3+\frac1{x^3}, \dots , x^n+\frac1{x^n}$$ is also a member of $\mathbb{N}$.  I have my base, it is indeed true for $n=1$..  I can assume it is true for $x^k+x^{-k}$ and then proove it is true for $x^{k+1}+x^{-(k+1)}$ but I'm stuck there.
26124
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:  $$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right)  }{ 2 } .$$  If $n=1$  Then  $$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$  Now assume $n=k$.  Thus  $$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$  If now $n=k+1$  $$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$  Is there anything more that I have to do, or is this it?
120543
Show:       $$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$   I showed this using the . I want other methods to prove it. Thank you.
99631
I have tried the following: $$ f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 $$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
49467
$$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$  I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.     My Attempt  This can be rewritten as   $$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$  Now consider   $$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$  Taking the derivative   $$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$  By symmetry we have  $$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$  Using W|A I got   $$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$  By integeration we have   $$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$  Let $x = 2/a$  $$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$  Question  I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
82445
Edit: the question had some missing details. It should read as follows:     Prove for all $a \in (0,\frac{\pi}{2}]$, $f_n \rightarrow f$ uniformly on $[a,\frac{\pi}{2}]$.  Here $$f_n(x) = \frac{n x}{1 + n \sin(x)}.$$   Here is my attempt at the problem:  If $x \in (0, \frac{\pi}{2}]$ then $| f_n(x) - f(x) |$  $$ =\left| \frac{nx}{1 + n\sin(x)} - \frac{x}{\sin(x)}\right| $$  $$ = \left|\frac{nx \sin(x) - x(1 + n \sin(x))}{[1+ n \sin(x)]\sin(x)}\right|$$  $$ = \frac{x}{\sin(x) + n \sin^2(x)} \leq \frac{1}{n} \space \space  \text{(is this line correct)}$$  So $\forall \epsilon &gt; 0$, we may choose $N \geq \frac{1}{\epsilon}$ such that when $n \geq N \implies |f_n(x)-f(x)| \leq \epsilon \space \space \forall x \in (0, \frac{\pi}{2}]$  It's already been established that  $f_n(x)$ converges pointwise to: $f(x) = 0, x =0$ and $f(x) = \frac{x}{\sin(x)} ,x \in (0, \frac{\pi}{2}]$
88391
The goal is to show (preferably without contour integration, as my knowledge is pretty limited there, but if you can do it that way there is no problem to share it) that: $$I= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}\mathrm dx=\frac{1}{240}$$  I am still trying to find more relevant approaches, but so far I thought of: Considering $$J(a)= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{a^2+\ln^2\left(\frac{1+x}{1-x}\right)}dx$$ We have that $J'(\pi)=-2\pi I$, also $$J(a)=\int_0^\infty e^{-at}\int_0^1 x\sin\left(t\ln\left(\frac{1+x}{1-x}\right)\right)dx dt $$ Which is nice, however I couldnt solve the inner integral.  Another approach would be to let $\ln\left(\frac{1+x}{1-x}\right)=t \rightarrow x=\frac{e^t-1}{e^t+1}\rightarrow dx=\frac{2e^t}{(e^t+1)^2}$  $$I=2\int_0^\infty \frac{e^t(e^t-1)}{(e^t+1)^3}\frac{t}{(\pi^2+t^2)^2}dt$$ This doesnt look nice, but notice that $$\frac{d^2}{dx^2} \left(\frac{1}{1+e^x}\right)=\frac{e^x(e^x-1)}{(e^x+1)^3}$$ Integrating by parts two times I get $$I=\frac{1}{\pi^2} +24 \int_0^\infty \frac{x^2-\pi^2}{(x^2+\pi^2)^4}\frac{x}{1+e^x}dx$$ Now I am thinking to use series for $\frac{x}{1+e^x}$, but I dont know a series that satisfies it, similary we have for $\frac{x}{e^x-1}$ in terms of Bernoulli numbers. Could you land some help for this integral?
68740
This one should be pretty easy as its only a first year math problem. it s proof by induction that Im having trouble with.  $$let\;x_{1}=1; for\;each\;n\in \mathbb{N}\;let\;x_{n+1}=\frac{2}{3}x_{n}+1; prove\; x_{n}&lt; 3\;for\;all\;n$$  If anyone is willing to spell it out really obviously step by step and explain how they got from one step to the next I would appreciate it a lot. Thanks.
25806
So for a better presentation:   I need to prove this using mathematical induction, so i started with a base of $n=1$  Since $i$ starts at $i = 1$ I got this:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{2} = \frac{1}{2}$$  So since this was true I went to the induction part: I chose $n = k$ and assumed this was true:  $$\frac{1}{i(i+1)} = \frac{n}{n+1}$$ $$\frac{1}{i(i+1)} = \frac{k}{k+1}$$  but I got stuck here since I didn't know if I needed to increment $i$ or not and because I didn't know if I needed to add the summation symbol or not and needed to do a few things with it.
96688
My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
3970
I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this .
9291
Possible Duplicate:          How can we integrate $$\int_0^\frac{\pi}2\frac{\sin^nx}{\sin^nx+\cos^nx}\,\mathrm dx , \,\,\,\,\,\,\,\,\, n\in N \quad?$$ Thanks for any hint.