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3

Assume, all the variables are real and positive so $a>0$: $$\int_{0}^{a}\frac{1}{\left(a^2+x^2\right)^{\frac{3}{2}}}\space\text{d}x=$$ Substitute $u=\arctan\left(\frac{x}{a}\right)$ and $\text{d}x=a\sec^2(u)\space\text{d}u$; This gives a new lower bound $u=\arctan\left(\frac{0}{a}\right)=0$ and upper bound ...


3

Reality Check: What is the integration variable for the inner integral doing in the outer bounds of the outer integral? $$\displaystyle\int_0^\infty\int_{-pt}^0 \textsf{stuff}\operatorname d x\operatorname d t \neq \int_{\color{red}{-pt}}^0 \int_0^\infty \textsf{stuff}\operatorname d t\operatorname d x $$ What we have is ...


2

I'm assuming $\ln$ is defined as the logarithm with base $e$, and that $e$ is defined as $\lim_{v\rightarrow 0}(1+v)^{1\over v}$; also, I'm sweeping a bunch of stuff (mainly proofs that the relevant functions are continuous, and that $e$ exists) under the rug in the interests of readability. It's a little easier (in my opinion) to turn it around, and ...


1

$\log(x)$ is frequently defined by that integral. The only other definition that you ever see is that it's the function $f(x)$ that satisfies $e^{f(x)}=f(e^{x})=x$. We can then use the fact that $\frac{d}{dx}e^x=e^x$ and the inverse fuction theorem to conclude that $$\frac{d}{dx}\ln(x)=\frac{1}{e^{\ln(x)}}=\frac{1}{x}$$


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The body on which you want to carry on the surface integral can be parametrized as $$\;r(t)=(\cos\theta,\,\sin\theta,\,t)\;,\;\;0\le \theta\le2\pi\;,\;\;0\le t\le 1+\cos\theta$$ and we thus get: $$r_t=(0,0,1)\;,\;\;r_\theta=(-\sin\theta,\,\cos\theta,\,0)\implies$$ $$\implies r_t\times ...


1

Differentiating we get $$ y'=1-4\int_0^ty(x)dx-4ty+4ty=1-4\int_0^ty(x)dx. $$ Differentiating again, we obtain $$ y''=-4y, $$ which has the solutions $y(t)=a\cos(2t)+b\sin(2t)$. Substituting in the original equation gives $$ a\cos(2t)+b\sin(2t)=t[1-2a\sin(2t)+2\cos(2t)-2]+4\int_0^t s(a\cos(2s)+b\sin(2s))ds. $$ Solving for $a$ and $b$, you finally compute ...


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With this kind of problem, the first thing is to get rid of the square root. So, just as John Barber commented, use $x=u^2$, $dx=2u\,du$. This makes $$I=\int\frac{1}{1+x^{1/2}}\,dx=2\int \frac{ u}{1+u}\,du=2\int \frac{ 1+u-1}{1+u}\,du=2\Big(\int du-\int\frac{ du}{1+u} \Big)$$ I am sure that you can take it from here.


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Let $u=x+A/x$ so that $x^2-u x+A=0$, or $$x = \frac{u}{2} \pm \frac12 \sqrt{u^2-4 A} $$ $$dx = \frac12 \left (1 \pm \frac{u}{\sqrt{u^2-4 A}} \right ) du$$ Note that the integration limits provided by the mapping $x \mapsto u$ depend on whether the point $x=B$ is less than or greater than the minimum of $u$ at $x=\sqrt{A}$. Let's assume the former, i.e., ...


1

Well, to follow the aproach you took. First of all, the usual parametrization of a line segment $[a,b]$ is $$\gamma(t)=(1-t)\cdot a+t\cdot b$$ when $t\in [0,1]$, so at the time $t=0$, you get the point $a$, and at the time $t=1$, you get the point $b$. Now, in your particular case, you need two parametrizations, one for the line segment $[0,1+i]$ and the ...


1

Following your approach, let $x = a\tan \theta$. Then $dx = a \sec^{2} \theta$ and using $\frac{1}{\sec \theta} = \cos \theta$ and $\tan^{2} \theta + 1 = \sec^{2} \theta$, we have: $I = \int_{0}^{a} \frac{1}{\left(a^{2} + x^{2} \right)^{\frac{3}{2}}} dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{a^{3}\sec^{3} \theta} \cdot a\sec^{2} d \theta = \frac{1}{a^{2}} ...


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The residue is simply the coefficient of $(z-1)^{-1}$, or in this case, $1/2! + 1/1! = 3/2$.


1

Use the change of variables $t= \sin(x) $ to get $$ I = \int_0^1 \frac{\ln^n t} {\sqrt{1- t^2}} dt$$ Then make another change of variables $t^2=u$ and simplify . Then see my answer to finish the problem.


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We have: $$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$ hence: $$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$


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Hint. By using the symmetry, the sought area is twice $$ \int_0^b(0.5x^2-2x+8)dx-\int_0^b(0.25x^2+x)dx $$ with the convenient solution of $$ 0.5b^2-2b+8=0.25b^2+b $$ that is $$b=4.$$


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Your are right, you only have to calculate one side (here the right side $x>0$) and then multiple the result by two. Your are interested in the are area between the graph $y=1/2x^2-2x+8$ and the graph $y=1/4x^2+x$. The are lies between $x=0$ and where the graphs hit each other at $$1/2x^2-2x+8=1/4x^2+x$$. This leads to $x=4$ respective $x=8$ but your ...


1

First off, $$\int \frac{1}{x}\,dx = \ln x +C$$ Second, using integration by parts, if you let $u = 1$, and $dv = \frac{1}{x}$, then $$uv-\int v\,du = \ln x - \int 0\,du =\ln x - K = \ln x +C$$ if I let $C = -K$. If you let $u = \frac{1}{x}$, and $dv = 1$, then $$uv-\int v\,du = \frac{1}{x}\cdot x - \int x\cdot \frac{-1}{x^2}\,dx = 1+\int\frac{1}{x}\,dx = ...



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