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## Hot answers tagged integration

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 ...

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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 ...

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$\log(x)$ 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 $$\frac{d}{dx}\ln(x)=\frac{1}{e^{\ln(x)}}=\frac{1}{x}$$

1

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. $$Solving for a and b, you finally compute ... 1 With this kind of problem, the first thing is to get rid of the square root. So, just as John Barber commented, use x=u^2, dx=2u\,du. This makes$$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)$$I am sure that you can take it from here. 1 Let u=x+A/x so that x^2-u x+A=0, or$$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$$when t\in [0,1], so at the time t=0, you get the point a, and at the time t=1, you get the point b. Now, in your particular case, you need two parametrizations, one for the line segment [0,1+i] and the ... 1 Following your approach, let x = a\tan \theta. Then dx = a \sec^{2} \theta and using \frac{1}{\sec \theta} = \cos \theta and \tan^{2} \theta + 1 = \sec^{2} \theta, we have: I = \int_{0}^{a} \frac{1}{\left(a^{2} + x^{2} \right)^{\frac{3}{2}}} dx = \int_{0}^{\frac{\pi}{4}} \frac{1}{a^{3}\sec^{3} \theta} \cdot a\sec^{2} d \theta = \frac{1}{a^{2}} ... 1 The residue is simply the coefficient of (z-1)^{-1}, or in this case, 1/2! + 1/1! = 3/2. 1 Use the change of variables t= \sin(x)  to get$$ I = \int_0^1 \frac{\ln^n t} {\sqrt{1- t^2}} dt$$Then make another change of variables t^2=u and simplify . Then see my answer to finish the problem. 1 We have:$$ \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} $$1 Hint. By using the symmetry, the sought area is twice$$ \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.$$1 Your are right, you only have to calculate one side (here the right side x>0) and then multiple the result by two. Your are interested in the are area between the graph y=1/2x^2-2x+8 and the graph y=1/4x^2+x. The are lies between x=0 and where the graphs hit each other at$$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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