User:Summa77

"Keep science and war apart; science gains, war degrades." ~Summa:77 $$<\begin{pmatrix} & &  & \\  &  &  & \\  &  &  & \\  &  &  & \end{pmatrix}$$ 1. $$\pi^2=\sum_{i=0}^{\infty} \frac{8}{(2i+1)^2}$$ 2. $$\pi^4=\sum_{i=0}^{\infty} \frac{96}{(2i+1)^4} $$ 3. $$dv=\frac{\partial s}{\partial t}$$ 4. $$da=\frac{{\partial}^2s}{\partial t^2}$$ 5. $$Area=\int_a^b f(x)dx$$ 6. $$Work=\int_{C} \vec{F}\cdot d\vec{r}$$ 7. $$Flux=\iint_{S} \vec{F}\cdot\hat{n} dS$$ 8. $$Mass=\iiint_{R} \rho dV$$ 9. $$mass=\rho\iint_{R} f(x,y)dA ,\forall \frac{\partial \rho }{\partial x^m}=0 ,\forall  m \in Z^+$$ 10. $$\mathcal{G_{\mu\nu}}=8\mathcal{\pi T_{\mu\nu}}$$ 11. $$\nabla_v T_{\mu\nu}=T_{\mu\nu,v}=\frac{\partial T_{\mu\nu}}{\partial x^v}+\Gamma^{r}_{v \mu} T_{\mu\nu}+\Gamma^{r}_{v\nu} T_{\mu\nu}$$ 12. $$\oint_{C} \vec{F}\cdot d{\vec{r}}=\iint_{R}\nabla\times\vec{F}dA$$ 13 $$p(n)=\sum_{i=1}^{n} \frac{n!}{(n-i)!i!} p^i q^{n-i}=\sum_{i=1}^{n} {}^{n}\textrm{C}_{i} p^{i}q^{n-i} $$ 14 $$\frac{{\partial}^{n}f}{\partial {x_1}^{n}}=\lim_{\Delta x_1\to0} \sum_{i=0}^{n} \frac{{}^{n}\textrm{C}_{i} (-1)^{n-i} f(a+i\Delta x_1, x_2,...x_n)}{\Delta {x_1}^n}$$ 15 $$\frac{1}{\pi}=\frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}$$ 16 $$\oint_{C} \vec{F}\cdot d\vec{r}=\iint_{S} (\nabla\times\vec{F})\cdot \hat{n}dS$$ $$\iiint_{D}\nabla\cdot\vec{F}dV=\iint_{S}\vec{F}\cdot\hat{n} dS$$ In the case of Euclidean geometry $$g_{\mu\nu}\leftrightarrow \begin{bmatrix} 1 &0 &0 \\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$$

Example 1
Let's compute: $$\int \frac{x^3}{\sin^2{x}+1}dx$$ Let's use integration by parts: $$\int udv=uv-\int vdu$$ Let: $$u=x^3$$ $$v=\frac{\tan^{-1}(\sqrt{2}\tan{x})}{\sqrt{2}}$$ $$du=3x^2dx$$ $$dv=\frac{1}{(\sin^2{x}+1)}dx$$ Now we can write: $$\frac{x^3\tan^{-1}(\sqrt{2}\tan{x})}{\sqrt{2}}-\frac{3}{\sqrt{2}}\int x^2\tan^{-1}(\sqrt{2}\tan{x})dx$$ This is about as far as we can get without having a whole heap of polylogarithmic functions of different orders.

Example 2
Let's compute the relatively easy integral: $$\int \frac{\cos^2{x}}{6(\cos^2{x}+\sin^2{x})-6+cot^3{x}}dx$$ In this case we can use the trigonometric identity: $$\sin^2{x}+\cos^2{x}=1$$ Therefore, we know that the denominator will simplify out to $$6(1)-6+\cot^3{x}=\cot^3{x}$$ Therefore the integral becomes: $$\int \frac{\cos^2{x}}{cot^3{x}}dx=\int \cos^2{x}\tan^3{x}dx=\int \sin^2{x}\tan{x}dx$$ Now we can use integration by parts to solve this: Integration by parts formula: $$\int udv=uv-\int vdu$$ Let: $$u=\sin^2{x}$$ $$du=\sin{2x}dx$$ $$dv=\tan{x}dx$$ $$v=-\ln{(\cos{x})}$$ Then: $$\int \sin^2{x}\tan{x}dx=-\sin^2{x}\ln{(\cos{x})}+\int \ln{(cos{x})}\sin{2x}dx$$ Let $$u=\cos{x}, du=-\sin{x}dx$$ $$=-\sin^2{x}\ln{\cos{x}}+\int \ln{(u)}udu$$ $$=u^2\ln{u}-u^2-\int u\ln{u}du-\int udu$$ $$=u^2\ln{u}-u^2-\frac{u^2}{2}-\int u\ln{u} du$$ $$=u^2\ln{u}-\frac{3u^2}{2}-\int u\ln{u}du$$ $$\int \ln{cos{x}} sin{2x}dx=$$

Example 3
Compute: $$\int \frac{1}{x^2(x^2+1)}dx$$ We can use integration by trigonometric substitution in this case: Let $$x=\tan{\theta}$$ $$dx=\sec^2{\theta}$$ $$\theta=\tan^{-1}{x}$$ $$\frac{1}{x^2}=\cot^2{\theta}$$ $$\frac{1}{1+x^2}=\cos^2{\theta}$$ $$\frac{1}{x^2(1+x^2)}=\cot^2{\theta}\cos^2{\theta}$$ $$dx=sec^2{\theta}d{\theta}$$ $$\int \cot^2{\theta}\cos^2{\theta}\sec^2{\theta}d{\theta}$$ $$=\int \cot^2{\theta} d{\theta}$$ $$=-\cot{\theta}-\theta+C$$ Substitute back: $$=-\cot\left ( {\tan^{-1}{x}} \right )-\tan^{-1}{x}+C$$ $$=-\frac{1}{x}-\tan^{-1}{x}+C$$