Evaluate the integral and name the antiderivative $g$

$\int \frac{1}{7\+2 \sin \left(x\right)\+3 \cos \left(x\right)} \mathit{\ⅆ}x$ = $\frac{1}{3}\arctan \left(\frac{1}{3}\+\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)$$\stackrel{\text{assign to a name}}{\to }$$g$$$

Calculate the jump at $x\=\mathrm{\π}$ 

$\underset{x\to \mathrm{\π}-}{lim}g-\underset{x\to \mathrm{\π}\+}{lim}g$ = $\frac{1}{3}\mathrm{\π}$$$

$Q≔{\int }_0^x\frac{1}{7\+2 \sin \left(t\right)\+3 \cos \left(t\right)} \mathit{\ⅆ}t\:$

$\mathrm{simplify}\left(\mathrm{value}\left(Q\right)\right) assuming x\>0$

$\&lcub;\begin{array}{cc}-\frac{1}{3}\arctan \left(\frac{1}{3}\right)\&plus;\frac{1}{3}\arctan \left(\frac{1}{3}\&plus;\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)& x\le \mathrm{\&pi;}\\ \frac{1}{3}\mathrm{\&pi;}\mathrm{floor}\left(-\frac{1}{2}\frac{\mathrm{\&pi;}-x}{\mathrm{\&pi;}}\right)-\frac{1}{3}\arctan \left(\frac{1}{3}\right)\&plus;\frac{1}{3}\arctan \left(\frac{1}{3}\&plus;\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)\&plus;\frac{1}{3}\mathrm{\&pi;}& \mathrm{\&pi;}<x\end{array}$

$\mathrm{simplify}\left(\mathrm{value}\left(Q\right)\right) assuming x<0$

$\&lcub;\begin{array}{cc}-\frac{1}{3}\arctan \left(\frac{1}{3}\right)\&plus;\frac{1}{3}\arctan \left(\frac{1}{3}\&plus;\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)\&plus;\frac{1}{3}\mathrm{ceil}\left(-\frac{1}{2}\frac{\mathrm{\&pi;}-x}{\mathrm{\&pi;}}\right)\mathrm{\&pi;}& x<-\mathrm{\&pi;}\\ -\frac{1}{3}\arctan \left(\frac{1}{3}\right)\&plus;\frac{1}{3}\arctan \left(\frac{1}{3}\&plus;\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)& -\mathrm{\&pi;}\le x\end{array}$

$Q≔\int \frac{1}{7\&plus;2 \sin \left(x\right)\&plus;3 \cos \left(x\right)} \mathit{\&DifferentialD;}x\&colon;$

$q_1≔\mathrm{IntegrationTools}:-\mathrm{Change}\left(Q\&comma;z\&equals;\tan \left(x\&sol;2\right)\&comma;z\right)$

$\&int;\frac{2}{\left(7\&plus;2\sin \left(2\arctan \left(z\right)\right)\&plus;3\cos \left(2\arctan \left(z\right)\right)\right)\left(z^2\&plus;1\right)}\&DifferentialD;z$

$q_2≔\mathrm{simplify}\left(\mathrm{expand}\left(q_1\right)\right)$

$\&int;\frac{1}{2z^2\&plus;2z\&plus;5}\&DifferentialD;z$

$q_3≔\mathrm{Student}:-\mathrm{Precalculus}:-\mathrm{CompleteSquare}\left(q_2\right)$

$\&int;\frac{1}{2{\left(z\&plus;\frac{1}{2}\right)}^2\&plus;\frac{9}{2}}\&DifferentialD;z$

$q_4≔\mathrm{value}\left(q_3\right)$

$\frac{1}{3}\arctan \left(\frac{2}{3}z\&plus;\frac{1}{3}\right)$

$\mathrm{eval}\left(q_4\&comma;z\&equals;\tan \left(x\&sol;2\right)\right)$

$\frac{1}{3}\arctan \left(\frac{1}{3}\&plus;\frac{2}{3}\tan \left(\frac{1}{2}x\right)\right)$

Table 6.6.5(b)   Maple calculation of antiderivative from first principles