2023-11-10

In [32]:

%%latex
\begin{equation}
\begin{aligned}
& \int \frac{x}{1 + \sin(x)} \, \mathrm{d}x  \\
= & \int\frac{x\cdot(1 - \sin(x))}{\cos^2(x)} \mathrm{d}x \\
= & \int x\cdot(1+\sin(x))\mathrm{d}(\tan(x)) \\
= & \int \arctan(y) \cdot (1+\frac{y}{\sqrt{1+y^2}}) \mathrm{d}y \\
= & y\cdot\arctan(y) - \frac 12 \log(1+y^2) - \log(y+\sqrt{y^2+1}) + \sqrt{1+y^2}\arctan(y) \\
= & tan(x)\cdot x - \log(\frac 
\end{aligned}
\end{equation}

%%latex \begin{equation} \begin{aligned} & \int \frac{x}{1 + \sin(x)} \, \mathrm{d}x \\ = & \int\frac{x\cdot(1 - \sin(x))}{\cos^2(x)} \mathrm{d}x \\ = & \int x\cdot(1+\sin(x))\mathrm{d}(\tan(x)) \\ = & \int \arctan(y) \cdot (1+\frac{y}{\sqrt{1+y^2}}) \mathrm{d}y \\ = & y\cdot\arctan(y) - \frac 12 \log(1+y^2) - \log(y+\sqrt{y^2+1}) + \sqrt{1+y^2}\arctan(y) \\ = & tan(x)\cdot x - \log(\frac \end{aligned} \end{equation}

\begin{equation} \begin{aligned} & \int \frac{x}{1 + \sin(x)} \, \mathrm{d}x \\ = & \int\frac{x\cdot(1 - \sin(x))}{\cos^2(x)} \mathrm{d}x \\ = & \int x\cdot(1+\sin(x))\mathrm{d}(\tan(x)) \\ = & \int \arctan(y) \cdot (1+\frac{y}{\sqrt{1+y^2}}) \mathrm{d}y \\ = & y\cdot\arctan(y) - \frac 12 \log(1+y^2) - \log(y+\sqrt{y^2+1}) + \sqrt{1+y^2}\arctan(y) \\ = & tan(x)\cdot x - \log(\frac \end{aligned} \end{equation}

In [4]:

%%latex
\begin{equation}
\frac 1 {2n+2} < \int_0^{\pi/4} \tan^n x dx < \frac 1 {2n}
\end{equation}

%%latex \begin{equation} \frac 1 {2n+2} < \int_0^{\pi/4} \tan^n x dx < \frac 1 {2n} \end{equation}

\begin{equation} \frac 1 {2n+2} < \int_0^{\pi/4} \tan^n x dx < \frac 1 {2n} \end{equation}

In [11]:

%%latex
\begin{equation}
\int_0^{1} e^x x dx = \left.(x-1)e^x\right\vert_{x=0}^{x=1} = 1 \neq e
\end{equation}

%%latex \begin{equation} \int_0^{1} e^x x dx = \left.(x-1)e^x\right\vert_{x=0}^{x=1} = 1 \neq e \end{equation}

\begin{equation} \int_0^{1} e^x x dx = \left.(x-1)e^x\right\vert_{x=0}^{x=1} = 1 \neq e \end{equation}

In [5]:

%%latex

\begin{equation}
\left\{
\begin{aligned}
&y''(x)=\frac{x^3}{y'(x)}+\frac{y'(x)}{x}, \\
&y(1)=0, \\
&y'(1)=2 \\
\end{aligned}
\right.
\end{equation}

%%latex \begin{equation} \left\{ \begin{aligned} &y''(x)=\frac{x^3}{y'(x)}+\frac{y'(x)}{x}, \\ &y(1)=0, \\ &y'(1)=2 \\ \end{aligned} \right. \end{equation}

\begin{equation} \left\{ \begin{aligned} &y''(x)=\frac{x^3}{y'(x)}+\frac{y'(x)}{x}, \\ &y(1)=0, \\ &y'(1)=2 \\ \end{aligned} \right. \end{equation}

In [6]:

%%latex
\begin{equation}
\lim\limits_{\lVert T\rVert\to 0}
\end{equation}

%%latex \begin{equation} \lim\limits_{\lVert T\rVert\to 0} \end{equation}

\begin{equation} \lim\limits_{\lVert T\rVert\to 0} \end{equation}

In [4]:

%%latex
\begin{equation}
\frac{1}{\sqrt{2}} \left(\log \left(\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}+1\right)-\log
   \left(1-\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}\right)\right)
\end{equation}

%%latex \begin{equation} \frac{1}{\sqrt{2}} \left(\log \left(\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}+1\right)-\log \left(1-\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}\right)\right) \end{equation}

\begin{equation} \frac{1}{\sqrt{2}} \left(\log \left(\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}+1\right)-\log \left(1-\frac{\sqrt{2} \sqrt{-x^2+x+2}}{x-2}\right)\right) \end{equation}

In [ ]: