Question Number 27612 by NECx last updated on 10/Jan/18
$$\int\frac{\mathrm{1}}{\mathrm{3}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by ajfour last updated on 11/Jan/18
$$=\int\frac{\mathrm{sec}\:^{\mathrm{2}} {xdx}}{\mathrm{3sec}\:^{\mathrm{2}} {x}+\mathrm{1}}=\int\frac{{d}\left(\mathrm{tan}\:{x}\right)}{\mathrm{4}+\mathrm{3tan}\:^{\mathrm{2}} {x}} \\ $$$${with}\:{t}=\mathrm{tan}\:{x}\:{it}\:{becomes} \\ $$$$\int\frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\left(\mathrm{2}/\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\mathrm{tan}\:{x}}{\mathrm{2}}\right)+{c}\:. \\ $$