# Tag Info

## New answers tagged integration

0

Note first that $$1+\cos x=2\cos^2\frac x2\tag{1}.$$ We can use (1) to perform the integration by parts because $\frac1{1+\cos x}$ is the derivative of $\tan\frac x2$: $$I=\int_0^{\pi/2}\frac{x+\sin x}{1+\cos x}\mathrm{d}x=\left[\left(\tan\frac x2 \right)\left(x+\sin x\right)\right]_0^{\pi/2}-\int_0^{\pi/2}\tan\frac x2(1+\cos x)\mathrm{d}x.$$ Using (1) ...

0

$\frac{d \log^2(2\sin(\pi x))}{dx}=2\log (2\sin(\pi x))\frac{d \log(2\sin(\pi x))}{dx}=-2 \pi \log (2\sin(\pi x)) \cot( \pi x)$

1

Using Chain rule:$$\frac{d }{dx}\log^2(2\sin(\pi x))=2\log(2\sin(\pi x))\times[2\times\pi\times\cos(\pi x)]\times\frac{1}{2\sin(\pi x)}$$

1

Let $f(x)=2\sin(\pi x)$. Then, $$\frac{d \log^2(f(x))}{dx}=2\log f(x)\times (\log f(x))^\prime.$$ Here, $$(\log f(x))^\prime=\frac{f^\prime (x)}{f(x)}=\frac{2\cos(\pi x)\times \pi}{2\sin(\pi x)}.$$ As a result, we have $$\frac{d \log^2(f(x))}{dx}=2\log (2\sin(\pi x))\times \frac{2\cos(\pi x)\times \pi}{2\sin(\pi x)}.$$ If you feel any difficulty in my ...

2

Hint: write the squared sum as a double sum. $$\left(\sum_{n=1}^\infty a_n\right)^2=\left(\sum_{n=1}^\infty a_n\right)\left(\sum_{m=1}^\infty a_m\right)=\sum_{m=1}^\infty\sum_{n=1}^\infty a_ma_n.$$

0

Have you tried the Divergence Theorem. In mathematics try to construct a visual or diagram of what you want to achieve before evaluating a complex integral. Draw these circles in 3D and use the divergence theorem or Green's Theorem.

0

Let $c_n$ be any sequence that tends to $\infty$. Since the sequence $(f_{c_n})$ converges in $L^2$, it is Cauchy in $L^2$. By Parseval's identity, the sequence $(\widehat{f_{c_n}})$ is Cauchy in $L^2$ as well: $$\|\widehat{f_{c_n}}-\widehat{f_{c_m}}\| = \|f_{c_n}-f_{c_m}\|$$ (maybe with some constant multiple, depending on the Fourier transform ...

1

Two quicker ways to do the same: Apply the residue theorem on the exterior domain $|z|>2$. The residue at $\infty$ is defined as $-c_{-1}$ where $c_{-1}$ is the coefficient of $z^{-1}$ in the Laurent expension about $\infty$. Since $zf(z)\to 0$ as $z\to\infty$, the residue is $0$. There are no poles in the exterior domain, either. Go from basic ...

7

Why not make the substitution $x=\sinh u?$ Then, $d x = \cosh u d u,$ while $(x+\sqrt{1+x^2})= \exp(u),$ and $1+x^2 = \cosh^2 u,$ so we get our integral equal to $$\int \frac{\sqrt u}{\cosh u} du,$$ Now, letting $u = v^2,$ this is equal to $$\frac12\int \operatorname{sech}(v^2) d v.$$ At which point I am flummoxed.

1

You need to be sure to use a unit normal. Since $\text{curl }F\cdot n$ is constant, you'll just get a constant times the area of the disk. No parametrization or explicit integration needed. By the way, it's easier to get the normal directly from the linear equation $a\cdot x=0$: We get $a=(1,2,2)$, and so $n=(1,2,2)/3$.

0

It may be easier to see ( at least it was for me) if you inscribe the cone is a unit- sphere since the height is equal to the radius, the triangle integrates over ${2\pi}$ much nicer. As for your trip integral approach... I am still working on that one myself.

1

$$x(\dfrac{1}{12}\tan(12x)-x)|_{0}^{\pi/36}-\int_{0}^{\pi/36}(\dfrac{1}{12}\tan(12x)-x)dx$$ $$x(\dfrac{1}{12}\tan(12x)-x)|_{0}^{\pi/36}-\frac{1}{12}\int_{0}^{\pi/36}\tan(12x)dx\color{red}{\underbrace{-}_{\text{sign error}}}\int_{0}^{\pi/36}xdx$$

2

Here's a lazy answer, that needs fact-checking (or a better answer): I wouldn't think there is a "brutal" proof because because the reason this is true when $M = \mathbb{R}^n$ and $N = \mathbb{R}^m$ is a property of the Lebesgue measure. So I would probably say: this is true when $M$ and $N$ are open sets in Euclidean spaces, so for the general case use a ...

1

We want, $$\left(\int_{0}^{4} \sqrt{x}\,\,dx-\int_{0}^{4}\dfrac12x\,\,dx\right)+\left(\int_{4}^{16}\dfrac12x\,\,dx-\int_{4}^{16}\sqrt{x}\,\,dx\right)$$ where the first term two terms give the area from $0$ to $4$ and the second term gives it from $4$ to $16$. The expression becomes: ...

2

Subbing $x=\arctan{u^2}$, $dx = (2 u)/(1+u^4) du$ produces $$2 \int du \frac{u^2}{1+u^4}$$ It turns out that $$1+u^4=(1+\sqrt{2} u+u^2)(1-\sqrt{2} u+u^2)$$ so that we may invoke partial fractions. The result is that the integral becomes $$\frac1{\sqrt{2}} \int du \left (\frac{u}{1-\sqrt{2} u+u^2} - \frac{u}{1+\sqrt{2} u+u^2} \right )$$ which is ...

1

$\sin^2{A}+\cos^2{A}=1$ implies $\tan^2{A}+1=\sec^2{A}$, so we let $x=R\tan{A}$ and dx=R\sec^2{A}dA$, and (x^2+R^2)^{3/2}=R^3\sec^3{A}$ so the integral becomes $$\int\frac{R^2\sec^2{A}}{R^3\sec^3{A}}.$$ Which is $\frac{1}{R}\int \cos{A}dA$. So it integrates as, $$\frac{\sin{A}}{R}.$$ Drawing a right triangle with tangent equal to $\frac{x}{R}$ we see it's ...

2

Substitute $x = R\tan \theta$ then $dx = R \sec^2 \theta d\theta$ and $x^2+R^2 = R^2(\tan^2\theta + 1) = R^2 \sec^2 \theta$. Note as well that $x = R\tan \theta$ implies that $\sin \theta = \dfrac{x}{\sqrt{x^2+R^2}}$. $$\int \frac{Rdx}{\left(x^2+R^2\right)^{3/2}} = \int \frac{R^2 \sec^2 \theta d\theta}{\left(R^2 \sec^2 \theta\right)^{3/2}} = \int ... 0 The difference shows up in operators which are not invariant under translation. For an invariant operator such as the Laplacian, the solution of \nabla^{2}L(x)=\delta(x) can be translated to give a solution of \nabla^{2}L_{y}(x)=\delta(x-y) by setting L_{y}(x)=L(x-y). But as soon as you add a potential such as 1/|x| to -\nabla^{2} (such as for an ... 0 Here is a sample from Maple:$$ \int \!\text{sech} ( x ) \text{sech} ( 1/3\,x ) \,dx= \\ \frac{2\left( 4\sqrt {3}\arctan \left( \frac {\sqrt {3}\sinh ( 1/3 x ) }{\cosh ( 1/3\,x ) } \right) ( \cosh ( 1/6 x ) )^2 - 2 \sqrt {3}\arctan \left( \frac {\sqrt{3} \sinh ( 1/3 x ) }{\cosh ( 1/ 3 x ) } \right) -3 \sinh ( 1/6\,x ) \cosh ( 1 /6 x ) \right) }{3\left( ...

5


1

Hint: Try $s = e^\frac{-u^2}{2}$, and $dv = \frac{1}{u^2}$

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