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-2

Integration by parts substitute $e^x+1=u$ $e^x dx = du$ it'll be simple by using Integration by parts


0

$$\int e^{2x} \sqrt{e^x+1}dx=\int (e^x\sqrt{e^x+1})e^xdx$$ $e^x+1=t,e^xdx=dt,e^x=t-1$ $$\int e^{2x}\sqrt{e^x+1}dx=\int(t-1)\sqrt tdt=\int t^{3/2}dt-\int t^{1/2}dt=$$ $$=\frac{t^{3/2+1}}{3/2+1}-\frac{t^{1/2+1}}{1/2+1}+C=\frac{2}{5}(e^x+1)^{5/2}-\frac{2}{3}(e^x+1)^{3/2}+C$$


1

HINT: $$\sqrt{e^x+1}=u\implies e^x+1=u^2$$


0

Rewrite the equation as $$y\frac{dy}{dx}=3x^2+4x^2.$$ Note that $$y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{y^2}{2}\right),$$ by the Chain Rule. So the derivative of $\frac{y^2}{2}$ is $3x^2+4x$. It follows that $$\frac{y^2}{2}=x^3+2x^2+C,$$ for some constant $C$. Use the value of $y$ when $x=1$ to evaluate $C$. Now we know everything.


0

$\dfrac{dy}{dx}=\dfrac{3x^2+4x}{y}$ Now cross-multiply to get: $y\,\mathrm{d}y=3x^2+4x\,\mathrm{d}x$. Integrate on both sides to get: $\frac{1}{2}y^2=x^3+2x^2+C$. Multiply by two on both sides to get: $y^2=2x^3+4x^2+C$. Now solve for $C$ by plugging in the coordinate pair $(1, \sqrt{10})$: $10=2\cdot1^3+4\cdot1^2+C$ gives $C=10-2-4=4$. $y^2$ becomes: ...


0

Rearranging: $$ydy=(3x^2+4x)dx$$ Integrate: $$\int ydy=\int(3x^2+4x)dx$$ $$\frac{y^2}2=x^3+2x^2+C$$ Put $(1,rad10)$ to get the value of C, then put $x=0$ to get y


0

$$\frac{\tan\left(\dfrac\pi4 + t\right) - \tan\dfrac\pi4}t =\frac{\sin\left(\dfrac\pi4 + t-\dfrac\pi4\right)}{t\cos\left(\dfrac\pi4 + t\right)\cos\dfrac\pi4}$$ $$\implies\lim_{t\to0}\frac{\tan\left(\dfrac\pi4 + t\right) - \tan\dfrac\pi4}t=\frac1{\cos\dfrac\pi4}\cdot\lim_{t\to0}\frac{\sin t}t\cdot\frac1{\lim_{t\to0}\cos\left(\dfrac\pi4 + t\right)}$$


0

A solution could be to consider Taylor expansion of $$\tan\left(\dfrac {\pi}4 + t\right)$$ built at $t=0$. You then have $$\tan\left(\dfrac {\pi}4 + t\right)=1+2 t+2 t^2+O\left(t^3\right)$$ and since $\tan(\dfrac {\pi}4)=1$, you have $$\frac{\tan\left(\dfrac {\pi}4 + t\right) - \tan\left(\dfrac{\pi}4\right)}t=\frac{1+2t+2t^2+\cdots-1}{t}=2+2t+\cdots$$ ...


1

The limit definition of the derivative is $f'(x) = \displaystyle\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$. Now, let $f(x) = \tan x$ and replace $x$ with $\frac{\pi}{4}$, $h$ with $t$. This gives us: $\displaystyle\lim_{t \to 0}\dfrac{\tan(\frac{\pi}{4}+t)-\tan(\frac{\pi}{4})}{t} = \dfrac{d}{dx}\left[\tan x\right]_{x = \frac{\pi}{4}}$. The AP Calc exam ...


0

This is the definition of derivative for $\tan(x)$ evaluated at $x=\frac{1}{4 \pi}$.


1

This function is derived by Slepian and Pollak in 1961 and this is also know by Prolate-spheroidal wave function. For more detail read these articles Prolate spheroidal wave functions, Fourier analysis, and uncertainty-I Prolate spheroidal wave functions, Fourier analysis, and uncertainty-II In 1964 J.F. kaiser introduce this function with simple ...


2

$\int (5^x+2e^{5 \ln x})dx=\int (e^{x\operatorname{ln}5}+2x^{5})dx=\frac{1}{\operatorname{ln}5}e^{x\operatorname{ln}5}+\frac13x^6+C$


0

A slightly different way to find this integral is as follows: $\displaystyle\int_{0}^{\frac{1}{2}}\frac{4}{4t^2+1}\; dt=\frac{4}{4}\int_{0}^{\frac{1}{2}}\frac{1}{t^2+\frac{1}{4}}\;dt=\frac{1}{\frac{1}{2}}\left[\arctan\frac{t}{\frac{1}{2}}\right]_{0}^{\frac{1}{2}}=2\left[\arctan 2t\right]_{0}^{\frac{1}{2}}$ ...


2

$$ \int_0^{\frac{1}{2}} \frac{4}{4t^2+1}\ dt $$ Let $u=2t$, $$ \frac{d}{dt}u=\frac{d}{dt}[2t]=2 \Rightarrow du = 2\ dt $$ $$ \int_0^1 \frac{2}{u^2+1}\ du= 2 \int_0^1 \frac{1}{u^2+1}\ du=2\arctan u\bigg|_0^1 $$ $$ =2\arctan 1 -2\arctan 0=2\frac{\pi}{4}-0=\frac{\pi}{2} $$


5

$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$ We set $t=\frac{\tan{u}}{2}$ we have the following: $t=0: u=0$ $t=\frac{1}{2}: u=\frac{\pi}{4}$ $dt=\frac{1}{2 \cos^2{u}}du$ $$\frac{4}{1+4t^2}=\frac{4}{1+4 \frac{\tan^2{u}}{4}}=\frac{4}{1+\tan^2{u}}=\frac{4 \cos^2{u}}{\sin^2{u}+\cos^2{u}}=4 \cos^2{u}$$ Therefore, we have the following: ...


2

$\int_0^{\infty}(x+1)e^{-x}dx=-(x+1)e^{-x}\Big|_0^{\infty}+\int_0^{\infty}e^{-x}dx=1+1=2$


1

Your answer is right. As for evaluating the integral, the limit at $\infty$ does exist, because $e^{-y}$ converges to zero as $y \rightarrow \infty.$ I think you should get an answer of $2$ after evaluating the integral, but you should check!


0

Let $x+1=\tan\theta$, so $x=\tan\theta-1$, $dx=\sec^{2}\theta d\theta$, and $\sqrt{x^2+2x+2}=\sqrt{(x+1)^2+1}=\sec\theta$. Then $\displaystyle\int\frac{x^2+4x}{\sqrt{x^2+2x+2}}dx=\int\frac{(\tan\theta-1)^2+4(\tan\theta-1)}{\sec\theta}\sec^{2}\theta d\theta$ $\displaystyle=\int(\tan^{2}\theta+2\tan\theta-3)\sec\theta ...


2

Most of the integrals I have found in my version of Gradshteyn and Rhyzik have as reference [BI], which means [BI] Bierens de Haan, D., Nouvelles tables d'integrales definies. Amsterdam, 1867. which can be found here https://archive.org/details/nouvetaintegral00haanrich The PDF has 762 scanned pages, which makes it really slow for viewing.


0

I've always had a terrible memory, so I just learned the double angle formula. $$\begin{array}{lll} \int\sqrt{1+\cos(t/2)}dt&=&\int\sqrt{1+\cos(2\cdot\frac{t}{4})}dt\\ &=&\int\sqrt{1+(\cos^2(t/4)-\sin^2(t/4))}dt\\ &=&\int\sqrt{1+\cos^2(t/4)-1+\cos^2(t/4)}dt\\ &=&\int\sqrt{2\cos^2(t/4)}dt\\ &=&... \end{array}$$ ...


0

$\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{\expo}[1]{\,{\rm e}^{#1}\,} ...


0

I would do it in two steps, and showing your work here is essential since the answer is $+\infty$. The first step is to evaluate the inner integral: $$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} F(x,y) \,dx \,dy = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \left\{\begin{array}{ccc}0 & : & 1 \le x+y \\ x+y & : & 0 < x+y < 1 ...


7

Okay, first of all: it is not easy to answer this question, and I'm not able to show the truth in an elegant way. I did the job with Maple computer algebra system (CAS). In this form there is no chance to evaulate to solution even with CAS, so we transform into the form, what @Lucian offered. Let $f$ be $$f:=\frac{\ln(\sinh x)}{1+ \sinh x} \cdot e^{x/2}$$ ...


4

This function is integrable in elementary terms. $$\int\frac1{\sqrt{3-5x-2x^2}}\,dx=\frac1{\sqrt2}\cdot\arcsin\left(\dfrac{4x+5}7\right)+C$$ It can be done with a combination of complete the square and rewriting the integral into an integral definition of $\arcsin x$. $$\int\frac1{\sqrt{1-x^2}}\,dx=\arcsin x+ C$$


3

A $k$-dimensional random vector $X=(X_n)_{1\leqslant n\leqslant k}$ is standard Gaussian if the entries $X_n$ are i.i.d. standard normal. The (Euclidean) length of $X$ is $L_k=\sqrt{X_1^2+\cdots+X_k^2}$, its expected length is $\lambda_k=E(L_k)$. Since the distribution of $X$ is completely known, one can compute $\lambda_k$ in the usual way, that is, by ...


8

A substitution works... $$ \frac{1}{(1+x\tan x)^2} = \frac{\cot^2 x}{(\cot x+x)^2} $$ and $$ \frac{d}{dx}\;(\cot x+x) = -\cot^2 x $$ so ... you finish it.


3

Consider the elliptic integral of the second kind $E(\phi,m):=\int\limits _0^\theta \sqrt{1-m \sin^2 x}\,dx$. (Note that my convention will be to write in terms of $m=k^2$ rather than $k$ itself.) Naively, the range of allowed $m$ is $m\in [0,1]$. The question posted above is then a special case of the following: What is the meaning of $E(\theta,-m)$ for ...


4

HINT: As $\displaystyle\frac{d(x^2+2x+2)}{dx}=2x+2,$ write $$x^2+4x=(x^2+2x+2)+(2x+2)-4$$ Use $\#1,\#8$ of this


9

Using IBP by setting $u=\ln^3(1+x)$ and $dv=\dfrac{\ln x}{x}\ dx$ yields \begin{align} \int_0^1\frac{\ln^3(1+x)\ln x}x\ dx&=-\frac32\int_0^1\frac{\ln^2(1+x)\ln^2 x}{1+x}\ dx\\ &=-\frac32\int_1^2\frac{\ln^2x\ln^2 (x-1)}{x}\ dx\quad\Rightarrow\quad\color{red}{x\mapsto1+x}\\ &=-\frac32\int_{\large\frac12}^1\left[\frac{\ln^2x\ln^2 ...


4

Is $\displaystyle\int\sqrt{1+\sin^2x}~dx~$ an elliptic integral? Yes, it is. In fact, that's how elliptic integrals came into being, by trying to compute the arc length of the sine or cosine function, which is precisely what you have done here. But when $k=i$? That seems a bit off. It does, doesn't it? But why is that such a shock for you? :-) ...


3

You may easily justify the following steps: $$ \begin{align} \int_0^1 x^{-x}\,dx& =\int_0^1 e^{-x\ln x}\,dx\\\\ &=\int_0^1 \sum_{k=0}^{\infty} \frac{(-x\ln x)^n }{n!}\,dx\\\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n }{n!}\int_0^1 (x\ln x)^n\,dx\\\\ &=\sum_{n=0}^{\infty} \frac{(-1)^n }{n!}(-1)^n \int_0^\infty u^ne^{-nu}\,du \quad ...


4

Hint: Make use of $x^{-x}=\exp(-x\ln x)=\sum_{k=0}^\infty (-1)^n\frac{(x\ln x)^n}{n!}$, now the integration of $n$th term corresponds to $(n+1)^{-n-1}$. By looking at $\int_0^1 x^n\ln^n{x}=(\partial_n)^n\int_0^1 x^ndx=(\partial_n)^n(\frac{1}{n+1})=(-1)^n n!\frac{1}{(n+1)^{n+1}}$ and plug in.


2

$$I=\int_{0}^{\pi^2/4}7\sin(\sqrt{x})dx$$ Let $u=\sqrt x$, $x=u^2$, $dx=2udu$ $$I'=14\int u\sin(u)du=14\left(u\int \sin u du-\int \left(\frac d{du}u\right)\left(\int \sin u du\right)du\right) \\=14(-u\cos u+\sin u)$$ $$I=14(-u\cos u+\sin u)_0^{\pi/2}=14$$


3

By the half-angle formulas, $$ 1+\cos(t/2) = 2\cos^2(t/4),$$ hence: $$\begin{eqnarray*}\int\sqrt{1+\cos(t/2)}\,dt&=&\sqrt{2}\int ...


1

Hint: Consider the complex valued function $f(x+iy):=P(x,y)+iQ(x,y)$. EDIT: For a multivariable calculus approach, observe that since $$\oint_C P(x,y)dx-Q(x,y)dy=0 $$ for all closed curves $C$, the differential 1-form $\omega:=Pdx-Qdy$ is said to be exact. Moreover, since exact forms are closed, we find $P_y=-Q_x$. Repeating the same reasoning gives the ...


3

$$ \begin{align} \int_0^\infty\frac{2\log(x)}{e^x+e^{-x}}\,\mathrm{d}x &=\frac{\partial}{\partial t}\int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x \end{align} $$ $$ \begin{align} \int_0^\infty\frac{2x^t}{e^x+e^{-x}}\,\mathrm{d}x &=\int_0^\infty2x^te^{-x}\left(1-e^{-2x}+e^{4x}-\dots\right)\,\mathrm{d}x\\ ...


0

Here is an interesting way we can go about this. It can be shown that (see this), if we replace $f(x)$ with $F$, and then replace $F^n$ with $f(x+nh)$, we have $$f^{(n)}(x)=\lim _{h\rightarrow0}\left( \frac{F-1}h\right)^n$$ Notice the neat fact that $\lim\limits_{h\to0}\frac{F-1}{h}=f'(x)$. For example, expanding the above for $n=2$ gives the limit formula ...


3

Just a partial answer for now. We have to compute: $$ I = \int_{0}^{+\infty}\frac{2\log x}{e^{x}+e^{-x}}\,dx = 2\frac{\partial}{\partial\alpha}\left.\left(\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz\,\right)\right|_{\alpha=0^+}$$ Since: $$\int_{0}^{+\infty}\frac{z^\alpha}{e^z+e^{-z}}\,dz=\int_{1}^{+\infty}\frac{\log^\alpha ...


0

Hint: try to interpret the three terms as the areas of three regions of the plane, delimited by the $x$ axis and the graph of $f$ or, for the middle one, by an opportune piecewise constant approximation of $f$. Then sketch those regions. They say a picture says more than a 1000 words.


0

$$\inf_{x\in [n, n+1]}f(x)\int_{n}^{n+1}dx\leq \int_{n}^{n+1}f(x)dx \leq \displaystyle\sup_{x\in [n, n+1]}f(x)\int_{n}^{n+1}dx$$ If $f$ is decreasing, then the left-most is equal to $f(n+1)$ and the right-most is equal to $f(n)$


1

Let $a,b$ be two integers then: $$\int_{a}^{b+1}f(x)dx=\sum_{k=a}^{b}\int_{k}^{k+1}f(x)dx\le\sum_{k=a}^{b}f(k)$$ since $f$ is decreasing (where I have used that $f(k)\ge f(x)$ for $x\in[k,k+1]$). Similarly $$\int_{a-1}^{b}f(x)dx=\sum_{k=a}^{b}\int_{k-1}^{k}f(x)dx\ge\sum_{k=a}^{b}f(k)$$


0

Alright, I'll try to answer it myself, now that the question has been discussed in the comments. Thanks a lot to Travis and André Nicolas! In general, we may always ignore the limits of integration. This is easily justifiable by the Fundamental Theorem of Calculus: $$\int_a^bf(x)\,dx = F(b) - F(a) = \left[ F(x) \right]_a^b = \left[ \int f(x)\,dx ...


4

Omran Kouba's answer is very good, but here is a perhaps simpler counterexample. Take $$f:[-2,2]\to\mathbb{R},\quad f(x)=x^2(x^2-12/5).$$ Then, $$\int_{-2}^2f(x)dx=f(0)=0,$$ but $$\int_{-2}^2f'(x)^2dx-2(f(-2)+f(2))^2=-\frac{19456}{175}<\frac{8}{(2-(-2))^2}\int_{-2}^2f(x)^2dx=\frac{47104}{7875}.$$


2

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1

Let $u=\ln x$ and $dv=\frac{1}{(x-3)^2} dx$, so $du=\frac{1}{x} dx$ and $v=-\frac{1}{x-3}$. Then $$\displaystyle\int\frac{\ln x}{(x-3)^2} \, dx=uv-\int v\,du=-\frac{\ln x}{x-3}-\int-\frac{1}{x(x-3)}\, dx=-\frac{\ln x}{x-3}+\int\frac{1}{x(x-3)}\, dx.$$ Now use partial fractions: $$\frac{1}{x(x-3)}=\frac{A}{x}+\frac{B}{x-3}\implies A(x-3)+Bx=1\implies ...


4

Note that $u$ is squared, not cubed, in the denominator. $$I = \int\frac{\ln(u+3)}{u^3}\mathrm{d}u$$ Using integration by parts, take $$w = \ln(u+3) \implies dw= \frac{1}{u+3}\,du$$ Then $$dv = \frac 1{u^2} \implies v = -\frac{1}{u}$$ $$I = wv - \int v\,dw$$ $$I = -\frac{\ln(u + 3)}{u} + \int \frac 1{u}\cdot \frac 1{u+3}\,du$$ Now you can use partial ...


1

$\int_0^\infty e^{-u}\dfrac{1}{\left(\sqrt{1+(h+u)^2}\right)^5}du$ $=\int_0^\infty\dfrac{e^{-u}}{(1+(u+h)^2)^\frac{5}{2}}du$ $=\int_h^\infty\dfrac{e^{-(u-h)}}{(1+u^2)^\frac{5}{2}}d(u-h)$ $=e^h\int_h^\infty\dfrac{e^{-u}}{(1+u^2)^\frac{5}{2}}du$ Consider $\int\dfrac{1}{(1+u^2)^\frac{5}{2}}du$ : Let $u=\tan\theta$ , Then $du=\sec^2\theta~d\theta$ ...


1

The following solution is based on a suggestion of lab bhattacharjee: $\displaystyle\int \frac{\sqrt{1-k^2\sin^2 x}}{\sin x} dx=\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin^{2} x}\cdot\sin x dx=\int\frac{\sqrt{1-k^2(1-\cos^{2}x)}}{1-\cos^{2}x}\cdot\sin x dx$. Now let $t=\cos x$, $dt=-\sin x dx$ to get $\displaystyle -\int\frac{\sqrt{1-k^2+k^2t^2}}{1-t^2}dt$; then ...


1

To prove the a.e. convergence, the method we think about may be the squeeze theorem. That is, we find two functions $F^*$ and $F_*$ that bounds the limit function. If the upper and lower bound are the same a.e., we can say that the limit exists a.e.. The natural way to choose the upper and lower bound, as we can see, is choosing $$F^*=\limsup F_n,F_* = ...


2

I am the author of the text referenced. The integral over infinite volume is indeed the delta function (times (2pi)^3) and seems in this context more straightforwardly expressed as e^2ipx. (There was a pedagogic reason for expressing it as two factors in (4-27).) In the text, the integral is actually over finite volume V and (4-27) has another factor in ...



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