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

$\=\frac{1}{\sqrt{13}}\int \frac{\mathrm{dx}}{\cos \left(x-\arctan \left(2\/3\right)\right)}$

$\=\frac{1}{\sqrt{13}}\int \sec \left(x-\arctan \left(2\/3\right)\right) \mathrm{dx}$

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

$2 \sin \left(x\right)\+3 \cos \left(x\right)\=A\left(\cos \left(x\right)\frac{3}{A}\+\sin \left(x\right)\frac{2}{A}\right)\Rightarrow \{\begin{array}{c}\cos \left(\mathrm{\phi }\right)\=3\/A\, and \sin \left(\mathrm{\phi }\right)\=2\/A\\ {\cos }^2\left(\mathrm{\phi }\right)\+{\sin }^2\left(\mathrm{\phi }\right)\=\frac{9\+4}{A^2}\=1\end{array}$ 

Consequently, $A^2\=13$ and $\mathrm{\φ}\=\arctan \left(2\/3\right)$

Table 6.6.4(a)   Establishing the identity $2 \sin \left(x\right)\+3 \cos \left(x\right)\=\sqrt{13}\cos \left(x-\arctan \left(2\/3\right)\right)$

$Q≔\int \frac{1}{2 \sin \left(x\right)\+3 \cos \left(x\right)} \ⅆx\:$

$q_1≔\mathrm{convert}\left(Q\,\mathrm{phaseamp}\,x\right)$

$\∫\frac{1}{13}\frac{\sqrt{13}}{\cos \left(x-\arctan \left(\frac{2}{3}\right)\right)}\ⅆx$

$q_2≔\mathrm{value}\left(q_1\right)$

$\frac{1}{13}\sqrt{13}\ln \left(\sec \left(x-\arctan \left(\frac{2}{3}\right)\right)\+\tan \left(x-\arctan \left(\frac{2}{3}\right)\right)\right)$

$\mathrm{eval}\left(q_2\,\ln \=\ln \@\mathrm{abs}\right)$

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

Maple's annotated stepwise solution

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$\int \frac{1}{2 \sin \left(x\right)\+3 \cos \left(x\right)} \ⅆx$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int \frac{1}{2\sin \left(x\right)+3\cos \left(x\right)}\ⅆx\\ \text{▫}& & \text{1. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=\tan \left(\frac{x}{2}\right)\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=2\arctan \left(u\right)\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{dx}=\frac{2\mathrm{du}}{u^2+1}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \frac{1}{2\sin \left(x\right)+3\cos \left(x\right)}\ⅆx=\int -\frac{2}{3u^2-4u-3}\ⅆu\\ & & \text{This gives:}\\ & & \int -\frac{2}{3u^2-4u-3}\ⅆu\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{2}{3u^2-4u-3}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{2}{3u^2-4u-3}\ⅆu=-2\left(\int \frac{1}{3u^2-4u-3}\ⅆu\right)\\ & & \text{We can rewrite the integral as:}\\ & & -2\left(\int \frac{1}{3u^2-4u-3}\ⅆu\right)\\ \text{▫}& & \text{3. Apply partial fraction decomposition}\\ & \text{◦}& \text{Partial fractions expansion}\\ & & \frac{1}{3u^2-4u-3}=-\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}-\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\\ & & \text{This gives:}\\ & & -2\left(\int \left(-\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}-\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\right)\ⅆu\right)\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \int \left(f\left(u\right)+g\left(u\right)\right)\ⅆu=\int f\left(u\right)\ⅆu+\int g\left(u\right)\ⅆu\\ & & f\left(u\right)=-\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}\\ & & g\left(u\right)=-\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\\ & & \text{This gives:}\\ & & -2\left(\int -\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}\ⅆu\right)-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{3\sqrt{13}}{26\left(-3u+2+\sqrt{13}\right)}\ⅆu=-\frac{3\sqrt{13}\left(\int \frac{1}{-3u+2+\sqrt{13}}\ⅆu\right)}{26}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{3\sqrt{13}\left(\int \frac{1}{-3u+2+\sqrt{13}}\ⅆu\right)}{13}-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{6. Apply a change of variables to rewrite the integral in terms of}\mathrm{u1}\\ & \text{◦}& \text{Let}\mathrm{u1}\text{be}\\ & & \mathrm{u1}=-3u+2+\sqrt{13}\\ & \text{◦}& \text{Isolate equation for}u\\ & & u=\frac{2}{3}+\frac{\sqrt{13}}{3}-\frac{\mathrm{u1}}{3}\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{du}=-\frac{\mathrm{du1}}{3}\\ & \text{◦}& \text{Substitute the values for u and du back into the original}\\ & & \int \frac{1}{-3u+2+\sqrt{13}}\ⅆu=\int -\frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\\ & & \text{This gives:}\\ & & \frac{3\sqrt{13}\left(\int -\frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{13}-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{7. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(\mathrm{u1}\right)\ⅆ\mathrm{u1}=C\left(\int f\left(\mathrm{u1}\right)\ⅆ\mathrm{u1}\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}=-\frac{\left(\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{3}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{\sqrt{13}\left(\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{13}-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{8. Apply the}\mathbf{\text{reciprocal}}\text{rule to the term}\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{reciprocal}}\text{rule}\\ & & \int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}=\ln \left(\mathrm{u1}\right)\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{\sqrt{13}\ln \left(\mathrm{u1}\right)}{13}-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{9. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}6\\ & & \mathrm{u1}=-3u+2+\sqrt{13}\\ & & \text{This gives:}\\ & & -\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}-2\left(\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\right)\\ \text{▫}& & \text{10. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{3\sqrt{13}}{26\left(3u-2+\sqrt{13}\right)}\ⅆu=-\frac{3\sqrt{13}\left(\int \frac{1}{3u-2+\sqrt{13}}\ⅆu\right)}{26}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}+\frac{3\sqrt{13}\left(\int \frac{1}{3u-2+\sqrt{13}}\ⅆu\right)}{13}\\ \text{▫}& & \text{11. Apply a change of variables to rewrite the integral in terms of}\mathrm{u1}\\ & \text{◦}& \text{Let}\mathrm{u1}\text{be}\\ & & \mathrm{u1}=3u-2+\sqrt{13}\\ & \text{◦}& \text{Isolate equation for}u\\ & & u=\frac{2}{3}-\frac{\sqrt{13}}{3}+\frac{\mathrm{u1}}{3}\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{du}=\frac{\mathrm{du1}}{3}\\ & \text{◦}& \text{Substitute the values for u and du back into the original}\\ & & \int \frac{1}{3u-2+\sqrt{13}}\ⅆu=\int \frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\\ & & \text{This gives:}\\ & & -\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}+\frac{3\sqrt{13}\left(\int \frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{13}\\ \text{▫}& & \text{12. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(\mathrm{u1}\right)\ⅆ\mathrm{u1}=C\left(\int f\left(\mathrm{u1}\right)\ⅆ\mathrm{u1}\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{1}{3\mathrm{u1}}\ⅆ\mathrm{u1}=\frac{\left(\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{3}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}+\frac{\sqrt{13}\left(\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\right)}{13}\\ \text{▫}& & \text{13. Apply the}\mathbf{\text{reciprocal}}\text{rule to the term}\int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{reciprocal}}\text{rule}\\ & & \int \frac{1}{\mathrm{u1}}\ⅆ\mathrm{u1}=\ln \left(\mathrm{u1}\right)\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}+\frac{\sqrt{13}\ln \left(\mathrm{u1}\right)}{13}\\ \text{▫}& & \text{14. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}11\\ & & \mathrm{u1}=3u-2+\sqrt{13}\\ & & \text{This gives:}\\ & & \frac{\sqrt{13}\ln \left(3u-2+\sqrt{13}\right)}{13}-\frac{\sqrt{13}\ln \left(-3u+2+\sqrt{13}\right)}{13}\\ \text{▫}& & \text{15. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}1\\ & & u=\tan \left(\frac{x}{2}\right)\\ & & \text{This gives:}\\ & & \frac{\left(\ln \left(3\tan \left(\frac{x}{2}\right)-2+\sqrt{13}\right)-\ln \left(-3\tan \left(\frac{x}{2}\right)+2+\sqrt{13}\right)\right)\sqrt{13}}{13}\end{array}$