User:The ipster

Using the substitution $$x=\pi-t$$, show that $$\int_0^\pi xf(\sin x) dx = \begin{matrix} \frac{\pi}{2} \end{matrix}\int_0^\pi f(\sin x) dx$$

$$\int_0^\pi \frac{x\sin x}{3+\sin^2 x} dx$$ $$\int_0^{2\pi} \frac{x\sin x}{3+\sin^2 x} dx$$ $$\int_0^\pi \frac{x|\sin x|}{3+\sin^2 x} dx$$

$$\lim_{x \to a} f(x) = +\infty \Leftrightarrow \forall\epsilon>0, \exists\delta>0, \forall x\in\mathbb{D}:0<|x-a|<\delta\Rightarrow f(x)<\epsilon$$

$$\left (\bigcap_{r=1}^n A_r\right)^{C}= \bigcup_{r=1}^n (A_r^{C})$$

$$\left (\bigcup_{r=1}^n A_r\right)^{C}= \bigcap_{r=1}^n (A_r^{C})$$

$$\Beta(p,q)=\int_0^1 x^{p-1} (1-x)^{q-1} dx$$ where $$p,q>0$$

$$\Gamma(k) = \int_0^\infty x^{k-1} e^{-x} dx$$ where $$k>0$$

$$\Beta(p,q) = 2\int_0^{\frac{\pi}{2}} \sin^{2p-1} y \cos^{2q-1} y dy$$

$$= \int_0^\infty \frac{y^{p-1}}{(1+y)^{p+q}} dy$$

$$\Beta(p,q) = \frac{\Gamma(p)\cdot\Gamma(q)}{\Gamma(p+q)}$$

$$\Gamma(1) = 1$$

$$\Gamma(2) = 1$$

$$\Gamma(3) = 2$$

$$\Gamma(n) = (n-1)!$$ where $$n\isin \mathbb{N}$$

$$\Gamma(k) = (k-1)\Gamma(k-1)$$ where $$k>1,k\isin \mathbb{R}$$

$$\Gamma(k)\cdot\Gamma(1-k) = \frac{\pi}{\sin k\pi}$$ where $$0<k<1$$

$$\Gamma(\begin{matrix} \frac{1}{2} \end{matrix}) = \sqrt\pi$$

$$\int_0^\frac{\pi}{2} \sin^m x \cos^n x dx = \frac{(m-1)(m-3)\ldots(n-1)(n-3)\ldots}{(m+n)(m+n-2)\ldots}\cdot K$$ where if m and n are both even then $$K = \begin{matrix} \frac{\pi}{2} \end{matrix}$$ otherwise $$K=1$$

$$\int_{-\infty}^{\infty} \frac{x}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma ^2}} dx = \mu$$

$$\int_{-\infty}^{\infty} \frac{(x-\mu)^2}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma ^2}} dx = \sigma ^ 2$$

$$F(s) = \bar f = \mathcal{L} \lbrace f(x) \rbrace = \int_0^\infty f(x) e^{-sx} dx$$

$$\mathcal{L} \lbrace 1 \rbrace = \frac{1}{s}$$

$$\mathcal{L} \lbrace x^k \rbrace = \frac {\Gamma(k+1)}{s^{k+1}}$$

$$\mathcal{L} \lbrace e^{kx} \rbrace = \frac{1}{s-k}$$

$$\mathcal{L} \lbrace \cos ax \rbrace = \frac{s}{s^2 + a^2}$$

$$\mathcal{L} \lbrace \sin ax \rbrace = \frac{a}{s^2 + a^2}$$

$$\mathcal{L} \lbrace f(x) + g(x) \rbrace = \bar f + \bar g$$

$$\mathcal{L} \lbrace e^{kx}f(x) \rbrace = F(s-k)$$

$$\mathcal{L} \lbrace f^\prime (x) \rbrace = s\bar f - f(0)$$

$$\mathcal{L} \lbrace f^n (x) \rbrace = s^n \bar f - s^{n-1}f(0) - s^{n-2}f^\prime (0) - \ldots - f^{(n-1)} (0)$$

$$\mathcal{L} \lbrace x^n f(x) \rbrace = (-1)^n \frac{d^n}{ds^n} (\bar f)$$ where $$n \isin \mathbb{N}$$

$$(f * g)(x) = \int_0^x f(t)g(x-t) dt$$

$$\mathcal{L} \lbrace f * g \rbrace = \mathcal{L} \lbrace f \rbrace \cdot \mathcal{L} \lbrace g \rbrace$$

$$\begin{pmatrix} \begin{matrix} \frac{\partial}{\partial x} \end{matrix}, \begin{matrix} \frac{\partial}{\partial y} \end{matrix}, \begin{matrix} \frac{\partial}{\partial z} \end{matrix} \end{pmatrix}$$

$$div(grad\phi ) = \nabla \cdot (\nabla \phi )= \nabla^2 \phi$$

$$\nabla_u \phi = \frac{\partial \phi}{\partial u} = \frac{\mathbf{u}}{|\mathbf{u}|} \cdot \nabla \phi$$

$$\iiint_{V} (\nabla \cdot \mathbf{v}) dV = \iint_{S} \mathbf{v} \cdot \mathbf{n} dS$$

$$\int_C \mathbf{v} \cdot d \mathbf{r} = \iint_S (\nabla \times \mathbf{v})\cdot \mathbf{n} dS$$