"Keep science and war apart; science gains, war degrades." ~Summa:77
${\displaystyle <{\begin{pmatrix}&&&\\&&&\\&&&\\&&&\end{pmatrix}}}$
1.
${\displaystyle \pi ^{2}=\sum _{i=0}^{\infty }{\frac {8}{(2i+1)^{2}}}}$
2.
${\displaystyle \pi ^{4}=\sum _{i=0}^{\infty }{\frac {96}{(2i+1)^{4}}}}$
3.
${\displaystyle dv={\frac {\partial s}{\partial t}}}$

4.
${\displaystyle da={\frac {{\partial }^{2}s}{\partial t^{2}}}}$
5.
${\displaystyle Area=\int _{a}^{b}f(x)dx}$
6.
${\displaystyle Work=\int _{C}{\vec {F}}\cdot d{\vec {r}}}$
7.
${\displaystyle Flux=\iint _{S}{\vec {F}}\cdot {\hat {n}}dS}$
8.
${\displaystyle Mass=\iiint _{R}\rho dV}$
9.
${\displaystyle mass=\rho \iint _{R}f(x,y)dA,\forall {\frac {\partial \rho }{\partial x^{m}}}=0,\forall m\in Z^{+}}$
10.
${\displaystyle {\mathcal {G_{\mu \nu }}}=8{\mathcal {\pi T_{\mu \nu }}}}$
11.
${\displaystyle \nabla _{v}T_{\mu \nu }=T_{\mu \nu ,v}={\frac {\partial T_{\mu \nu }}{\partial x^{v}}}+\Gamma _{v\mu }^{r}T_{\mu \nu }+\Gamma _{v\nu }^{r}T_{\mu \nu }}$
12.
${\displaystyle \oint _{C}{\vec {F}}\cdot d{\vec {r}}=\iint _{R}\nabla \times {\vec {F}}dA}$
13
${\displaystyle p(n)=\sum _{i=1}^{n}{\frac {n!}{(n-i)!i!}}p^{i}q^{n-i}=\sum _{i=1}^{n}{}^{n}{\textrm {C}}_{i}p^{i}q^{n-i}}$
14
${\displaystyle {\frac {{\partial }^{n}f}{\partial {x_{1}}^{n}}}=\lim _{\Delta x_{1}\to 0}\sum _{i=0}^{n}{\frac {{}^{n}{\textrm {C}}_{i}(-1)^{n-i}f(a+i\Delta x_{1},x_{2},...x_{n})}{\Delta {x_{1}}^{n}}}}$
15
${\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}}$
16
${\displaystyle \oint _{C}{\vec {F}}\cdot d{\vec {r}}=\iint _{S}(\nabla \times {\vec {F}})\cdot {\hat {n}}dS}$

${\displaystyle \iiint _{D}\nabla \cdot {\vec {F}}dV=\iint _{S}{\vec {F}}\cdot {\hat {n}}dS}$
In the case of Euclidean geometry ${\displaystyle g_{\mu \nu }\leftrightarrow {\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}}}$

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

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

Compute:

${\displaystyle \int {\frac {1}{x^{2}(x^{2}+1)}}dx}$
We can use integration by trigonometric substitution in this case:
Let ${\displaystyle x=\tan {\theta }}$

  ${\displaystyle dx=\sec ^{2}{\theta }}$

  ${\displaystyle \theta =\tan ^{-1}{x}}$

${\displaystyle {\frac {1}{x^{2}}}=\cot ^{2}{\theta }}$
${\displaystyle {\frac {1}{1+x^{2}}}=\cos ^{2}{\theta }}$
${\displaystyle {\frac {1}{x^{2}(1+x^{2})}}=\cot ^{2}{\theta }\cos ^{2}{\theta }}$
${\displaystyle dx=sec^{2}{\theta }d{\theta }}$
${\displaystyle \int \cot ^{2}{\theta }\cos ^{2}{\theta }\sec ^{2}{\theta }d{\theta }}$
${\displaystyle =\int \cot ^{2}{\theta }d{\theta }}$
${\displaystyle =-\cot {\theta }-\theta +C}$
Substitute back:
${\displaystyle =-\cot \left({\tan ^{-1}{x}}\right)-\tan ^{-1}{x}+C}$
${\displaystyle =-{\frac {1}{x}}-\tan ^{-1}{x}+C}$