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12

We have $$a_n=\int_0^1\frac{nx^{n-1}}{1+x}\,dx=\frac{x^n}{1+x}\Big|_0^1+\int_0^1\frac{x^n}{(1+x)^2}\,dx=\frac12+\int_0^1\frac{x^n}{(1+x)^2}\,dx \quad \forall n \ge 1.$$ Since $$\int_0^1\frac{x^n}{(1+x)^2}\,dx\le \int_0^1x^n\,dx=\frac{1}{n+1} \quad \forall n\ge 1,$$ it follows that $$\lim_n\int_0^1\frac{x^n}{(1+x)^2}\,dx=0.$$ Thus $\lim_na_n=\frac12$.

10

EDIT: I feel kind of stupid for not thinking of the easier ways in other posts, but I think this method is kind of cool. I apologize in advance, this is a lot of math and few words. \begin{align} \int_0^1\frac{nx^{n-1}}{1+x}\text dx&=\int_0^1nx^{n-1}\sum_{k=0}^\infty(-x)^k\text dx\\ &=\sum_{k=0}^\infty \int_0^1nx^{n-1+k}(-1)^k\text dx\\ ... 10 I can at least do the first one:\begin{align}\int_0^{\frac12}x\cot(\pi x)\text dx&=\left.\frac1\pi x\log(\sin(\pi x))\right|_0^{\frac12}-\int_0^{\frac12}\frac1\pi\log\sin(\pi x)\text dx\\ &=-\frac1{\pi^2}\int_0^{\pi/2}\log\sin x\ \text dx\tag 1\\ &=\frac{\log 2}{2\pi}\tag 2 \end{align}$$Justification for (1):$$\begin{align} \lim_{x\to 0} ...

9

Your integral can be related to derivatives of the Hurwitz Zeta function. But, an interesting tidbit is that Ramanujan derived the formula: $\displaystyle 1/2\int_{0}^{x}u^{n}\cot(u/2)du$ $\displaystyle=\cos(\frac{\pi n}{2})n!\zeta(n+1)-\sum_{k=0}^{n}(-1)^{\frac{k(k+1)}{2}}\frac{\Gamma(n+1)}{\Gamma(n+1-k)}x^{n-k}Cl_{k+1}(x)$ Where $Cl_{k+1}(x)$ is the ...

6

It's actually a lot simpler than this. Rewrite the integral as $$\int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\cosh^2{\delta}-\cosh^2{u}}} = \int_0^{\delta} du \frac{\cosh{u}}{\sqrt{\sinh^2{\delta}-\sinh^2{u}}}$$ Sub $y=\sinh{u}$ and the integral becomes $$\int_0^{\sinh{\delta}} \frac{dy}{\sqrt{\sinh^2{\delta}-y^2}}$$ Now sub $y=\sinh{\delta}\, \sin{t}$ and ...

5

These are both immediate using two subsitutions. Everything is even, so split all of them at $0$. \begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned} And: \begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}

4

Thinking about the graph of $x^n$ on $[0,1]$ we observe that it stays near 0 and then sharply jumps to 1. As such, it makes sense to break up the integral into $[0,c)$ and $[c,1]$ (for some $c$ to be chosen later). $$a_n = \int_0^c{\frac{n x^{n-1}}{x+1}dx} + \int_c^1{\frac{n x^{n-1}}{x+1}dx} \leq \int_0^c{n x^{n-1}dx} + \int_c^1{\frac{n x^{n-1}}{c+1}dx}\\ ... 4 I could have sworn when I began my answer there was a square in the integrand and not a 6 power on the 2x-1 term. Anyway, this is for (2x-1)^{2}. But, the same idea can be used on the 6 power, but it will be very long and messy. It may even make tech grunt and groan. One way to go about it is to use the identity: \displaystyle ... 3 \newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 ... 3 WARNING: This is the answer for the previous version of the question After substittion s=\frac{\pi}{2}-\pi x we get that our integral equals to$$ I=\frac{4}{\pi^3}\int_0^{\pi/2} s^2\log^2(2\cos s)ds $$From this answer it follows that$$ I=\frac{4}{\pi^3}\frac{11\pi^5}{1440}=\frac{11\pi^2}{360} $$3 [Initial apology: this is just a bunch of considerations too long to fit in a comment] Now we have a proof.$$ I = -\int_{0}^{+\infty} x\cdot\log^2(1-e^{-x})\,dx = \int_{0}^{+\infty}\frac{x^2}{e^{x}-1}\log(1-e^{-x})\,dx, I = -\int_{0}^{+\infty}\frac{x^2}{e^{x}-1}\sum_{k=1}^{+\infty}\frac{e^{-kx}}{k}dx = -\int_{0}^{+\infty}x^2 ...

2

Continuing from your substitution: \begin{aligned}\mathcal{I} &= \int_1^{\gamma} \frac{\gamma\,dv}{v\sqrt{(\gamma^2-v^2)(v^2-1)}}\\&=\int_0^{\frac{\pi}{2}}\frac{\gamma\,\sec^2 t dt}{1 +\gamma^2\tan^2 t}\quad (v^2=\cos^2 t+\gamma^2\sin^2 t)\\&=\int_0^{\infty}\frac{dw}{1 +w^2}\quad (w=\gamma\tan t)\\&=\frac{\pi}{2}\end{aligned}

2

Here is a start. Making the change of variables $u =\sin(\pi x)$ gives $$\frac{1}{\pi} \int _{0}^{1}\!{\frac { \ln^2\left( u/2\right) \left( 2\,{u}^{2}-1 \right) }{\sqrt {1-{u}^{2}}}}{du}.$$ Now, to evaluate the integral, you can follow the technique and the related links. You can do the second one using the same substitution and technique.

2

I know a way that may work here. I am assuming the integrals is regarded as $$2\int_0^{\infty}\frac{dx}{1+x^4}$$ First, show that $$\int_0^{\infty}\frac{x^{p-1}}{1+x}dx=\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(\pi p)},~~~0<p<1$$ This can be done by a especial substitution $y=\frac{x}{1+x}$. Second, after showing the above identity, then focus on ...

2

You have to change the variables because integration in the original variables is very difficult (i'm not sure how to do it), but changing to polar removes a variable because of the identity $\sin^2{\theta}+\cos^2{\theta}=1$, which you did. Also, because of the change of variable, you have to calculate the Jacobian which produces an additional term in the ...

2

Let $x = 1 -u^3$, we can rewrite the integral $\mathcal{I}$ as $$\mathcal{I} = \int_0^2\sqrt[3]{1-x^2}dx = 3\sqrt[3]{2}\int_{-1}^1 u^3 \left(1 - \frac{u^3}{2}\right)^{\frac13} du\\$$ Since the power series expansion of $\left(1 - \frac{u^3}{2}\right)^{\frac13}$ at $u = 0$ has radius of convergence $> 1$, we can expand it inside the integral sign and ...

2

You don't need calculus for the volume of something like this if you take Cavalieri's principle and dimensional rescaling as givens. (If you do not, see below for my comments on that.) Start be decomposing a $1\times1\times1$ cube into six equal square-based pyramids, using a vertex at the center of the cube. So the volume of these pyramids are each ...

1

The distance from the axis of rotation to the curve is $r = 2\sqrt a - \sqrt x$. So a vertical slice of the solid of revolution, of thickness $dx$, will have area $\pi r^2dx = \pi(4a + x - 4\sqrt{ax})dx$. You just have to integrate this expression (over the interval $[0,4a]$, if that is what you want).

1

You did $dx/dt = 2t - 3t^2$ and $dy/dt = 1 + 4t^3$ which leads us to have $$\frac{dy}{dx}=\frac{1 + 4t^3}{2t - 3t^2}$$ Now you were using this formula: $$P=\int_a^b2\pi y \sqrt{1+y'^2}dy$$ But what is $\sqrt{y'^2+1}$. $$\sqrt{y'^2+1}=\sqrt{1+\left(\frac{1 + 4t^3}{2t - 3t^2}\right)^2}=\frac{\sqrt{(2t-3t^2)^2+(1+4t^3)^2}}{\color{red}{|2t-3t^2|}}$$ Now you ...

1

Let $\displaystyle\arcsin x=\phi\implies\sin\phi=x$ and $-\frac\pi2\le \phi\le\frac\pi2$ based the principal value of inverse sine ratio So, $\displaystyle\cos(\arcsin x)=\cos\phi=+\sqrt{1-\sin^2\phi}=+\sqrt{1-x^2}$

1

$$\int\limits_{-\pi}^0e^{-x}\cos(nx)\,\mathrm{d}x=-\int\limits_{-\pi}^0(e^{-x})'\cos(nx)\,\mathrm{d}x =-e^{-0}\cos(n0)+e^{-\pi}\cos(n\pi)-n\int\limits_{-\pi}^0e^{-x}\sin(nx)\,\mathrm{d}x= -1+(-1)^ne^{-\pi}+n\int\limits_{-\pi}^0(e^{-x})'\sin(nx)\,\mathrm{d}x=$$ ...

1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 ... 1 Let$x = \sin \theta, \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx = 2\int_{0}^1 \frac{1}{\sqrt{1-x^2}} dx = 2 \int_0^{\pi/2} \frac{1}{\cos\theta} d\sin\theta =\pi\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx = \int_{-\infty}^{\infty} d \left(\tan^{-1}x \right) =\int_{-\pi/2}^{\pi/2} dy = \pi, where \quad y=\tan^{-1}x$1 If in the third integral you let$x=\tan(t)$, you get $$I_3 = \int_{-\pi/2}^{\pi/2} 1 dt.$$ If, in the second integral, you do the substitution$x=\sin(u)$, you get $$I_2 = 2\int_{-\pi/2}^{\pi/2} \cos^2 t ~ dt.$$ Alternatively, substitute$x = \cos(u)$to get $$I_2 = 2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt.$$ So combining these last two, $$I_2 = ... 1 This particular integral definitely looks simple. What changes when we try to evaluate it? To start with, a simple substitution like u=e^x, du=e^xdx goes like this:$$\int\frac{dx}{1+2^x+3^x}=\int\frac {e^xdx}{e^x+e^{x(\ln 2+1)}+e^{x(\ln 3+1)}}=\int\frac{du}{u+u^{\ln 2+1}+u^{\ln 3+1}}$$Already we have left the carefree world of simple functions ... 1 Define I_n =\displaystyle \int_0^1 \frac{x^n}{1+x}. Then you can obtain immediately that I_{n+1}+I_n = \displaystyle \frac{1}{n+1}. Next note that 0\leq I_{n+1}\leq I_n since for 0\leq x \leq 1 the inequality 0\leq \frac{x^{n+1}}{1+x} \leq \frac{x^n}{1+x} holds. Therefore I_n \to 0 as n \to \infty. Thus we have$$ a_{n+1}+a_n = ... 1 Consider the integrals $$I=\int_a^b f(x)\;\mathrm dx,\qquad J=\int_c^d g(y)\;\mathrm dy,$$ and assume that$f\gt0$everywhere and that$I=J$. Then the following change of variable transforms$I$into$J$. Consider the primitives$F$and$G$of$f$and$g$defined by $$F(x)=\int_a^x f(t)\;\mathrm dt,\qquad G(y)=\int_c^y g(s)\;\mathrm ds,$$ and the change ... 1 1) Evaluation of$I_{\max}(k,n)$It is clear that$\max x = x$so$I_{\max}(1,n)=\int\limits_0^1 x_1^n\,dx_1=\frac{1}{n+1}$. Also it is obvious that$\max\limits_{1\le i\le k}x_i=\max\left(x_k,\max\limits_{1\le i\le k-1}x_i\right)$. Then ... 1$\delta(x)$is not really a function in classical sense. For the purpose of deriving an expression without involving the concept of distribution, we will treat it as some sort of derivative of a step function. Assume all$k_i \ne 0$, let$$\lambda_i = |k_i|,\quad y_i = \begin{cases}x_i,& k_i > 0\\1-x_i,& k_i < 0\end{cases}, \quad K = ... 1$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 ...

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