$\int \frac{1}{2 \sin \left(x\right)\+3 \cos \left(x\right)} \mathit{\ⅆ}x$

$\= \int \frac{\frac{2 \mathrm{dz}}{1\+z^2}}{2\left(\frac{2 z}{1\+z^2}\right)\+3\left(\frac{1-z^2}{1\+z^2}\right)}$

$\= -2\int \frac{\mathrm{dz}}{3z^2-4 z-3}$

$\= -2\int \frac{\mathrm{dz}}{3{\left(z-2\/3\right)}^2-13\/3}$

$\=\frac{1}{\sqrt{13}}\int \left(\frac{1}{\left(z-2\/3\right)\+\sqrt{13}\/3}-\frac{1}{\left(z-2\/3\right)-\sqrt{13}\/3}\right) \mathit{\ⅆ}z$

$\=\frac{1}{\sqrt{13}}\ln \left(\left|z-2\/3\+\sqrt{13}\/3\right|\right)$ $-\frac{1}{\sqrt{13}} \ln \left(\left|z-2\/3-\sqrt{13}\/3\right|\right)$

$\=\frac{1}{\sqrt{13}}\ln \(\left|\frac{3 z-2\+\sqrt{13}}{3 z-2-\sqrt{13}}\right|\)$

$\=\frac{1}{\sqrt{13}}\ln \(\left|\frac{3 \tan \left(x\/2\right)-2\+\sqrt{13}}{3 \tan \left(x\/2\right)-2-\sqrt{13}}\right|\)$

$\int \frac{1}{2 \sin \left(x\right)\+3 \cos \left(x\right)} \mathit{\ⅆ}x$

$\= -2\int \frac{\mathrm{dz}}{3{\left(z-2\/3\right)}^2-13\/3}$

$\=\frac{6}{13}\int \frac{\mathrm{dz}}{1-\frac{9}{13}{\left(z-\frac{2}{3}\right)}^2}$

$\&equals;\&lcub;\begin{array}{cc}\frac{6}{13}\mathrm{arctanh}\left(\frac{3}{\sqrt{13}}\left(z-\frac{2}{3}\right)\right)\&comma;& \frac{3}{\sqrt{13}}\left|z-\frac{2}{3}\right|<1\\ \frac{6}{13}\mathrm{arccoth}\left(\frac{3}{\sqrt{13}}\left(z-\frac{2}{3}\right)\right)\&comma;& \frac{3}{\sqrt{13}}\left|z-\frac{2}{3}\right|\&gt;1\end{array}$

Figure 6.6.3(a)   $\mathrm{\&lambda;}<1$ in black; $\mathrm{\&lambda;}\&gt;1$ in red

Evaluate the integral

$\int \frac{1}{2 \sin \left(x\right)\&plus;3 \cos \left(x\right)} \mathit{\&DifferentialD;}x$ = $\frac{2}{13}\sqrt{13}\mathrm{arctanh}\left(\frac{1}{26}\left(6\tan \left(\frac{1}{2}x\right)-4\right)\sqrt{13}\right)$$$

$\begin{array}{c}\int \frac{1}{2\sin \left(x\right)+3\cos \left(x\right)}\mathit{\&DifferentialD;}x\hfill \\ & \text{=}& -2\int \frac{1}{3u^2-4u-3}\mathit{\&DifferentialD;}u\hfill & \left[\mathrm{change}\&comma;u=\tan \left(\frac{x}{2}\right)\&comma;u\right]\\ & \text{=}& \frac{3\sqrt{13}\left(\int \frac{1}{3u-2+\sqrt{13}}\mathit{\&DifferentialD;}u\right)}{13}+\frac{3\sqrt{13}\left(\int \frac{1}{-3u+2+\sqrt{13}}\mathit{\&DifferentialD;}u\right)}{13}\hfill & \left[\mathrm{partialfractions}\right]\end{array}$

Table 6.6.3(a)   Initial steps taken by the Integration Methods tutor

Initialize

$\mathrm{with}\left(\mathrm{IntegrationTools}\right)\&colon;$

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

Apply the rationalizing substitution $z\&equals;\tan \left(x\&sol;2\right)$ 

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

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

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

$-2\left(\&int;\frac{1}{3z^2-4z-3}\&DifferentialD;z\right)$

Partial-fraction decomposition of the integrand

$q_3≔\mathrm{allvalues}\left(\mathrm{convert}\left(\mathrm{GetIntegrand}\left(q_2\right)\&comma;\mathrm{fullparfrac}\&comma;z\right)\right)$

$\frac{1}{26}\frac{\sqrt{13}}{z-\frac{2}{3}-\frac{1}{3}\sqrt{13}}-\frac{1}{26}\frac{\sqrt{13}}{z-\frac{2}{3}\&plus;\frac{1}{3}\sqrt{13}}$

Integrate the partial fractions

$q_4≔\mathrm{Expand}\left(-2 \mathrm{Int}\left(q_3\&comma;z\right)\right)$

$-\frac{1}{13}\sqrt{13}\left(\&int;\frac{1}{z-\frac{2}{3}-\frac{1}{3}\sqrt{13}}\&DifferentialD;z\right)\&plus;\frac{1}{13}\sqrt{13}\left(\&int;\frac{1}{z-\frac{2}{3}\&plus;\frac{1}{3}\sqrt{13}}\&DifferentialD;z\right)$

${q_5}_{}≔\mathrm{value}\left(q_4\right)$

$-\frac{1}{13}\sqrt{13}\ln \left(z-\frac{2}{3}-\frac{1}{3}\sqrt{13}\right)\&plus;\frac{1}{13}\sqrt{13}\ln \left(z-\frac{2}{3}\&plus;\frac{1}{3}\sqrt{13}\right)$

Use "brute-force" to take the absolute value of the argument of the logarithms

$q_6≔\mathrm{eval}\left(q_5\&comma;\ln \&equals;\ln \&commat;\mathrm{abs}\right)$

$-\frac{1}{13}\sqrt{13}\ln \left(\left|z-\frac{2}{3}-\frac{1}{3}\sqrt{13}\right|\right)\&plus;\frac{1}{13}\sqrt{13}\ln \left(\left|z-\frac{2}{3}\&plus;\frac{1}{3}\sqrt{13}\right|\right)$

$\mathrm{combine}\left(q_6\&comma;\ln \right)$

$\sqrt{13}\ln \left(\frac{{\left|z-\frac{2}{3}\&plus;\frac{1}{3}\sqrt{13}\right|}^{1\&sol;13}}{{\left|z-\frac{2}{3}-\frac{1}{3}\sqrt{13}\right|}^{1\&sol;13}}\right)$

Table 6.6.3(b)   Stepwise calculation of a solution given in terms of logarithms of absolute values.