Integration Questions

Question Number 108507 by Eric002 last updated on 17/Aug/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \sqrt{\frac{\mathrm{3}{cos}\mathrm{2}{x}−\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}}\:{dx} \\$$

Answered by Sarah85 last updated on 18/Aug/20

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\sqrt{\frac{\mathrm{3cos}\:\mathrm{2}{x}\:−\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}}{dx}=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\sqrt{\frac{\mathrm{3cos}\:\mathrm{2}{x}\:−\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}}{dx} \\$$$$\mathrm{first}\:\mathrm{step} \\$$$${t}=\mathrm{tan}\:{x}\:\Leftrightarrow\:{x}=\mathrm{arctan}\:{t}\:\Leftrightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\$$$$\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}} {\int}}\frac{\sqrt{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\$$$$\mathrm{second}\:\mathrm{step} \\$$$${t}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:{u}\:\Leftrightarrow\:{u}=\mathrm{arcsin}\:\sqrt{\mathrm{2}}{t}\:\Leftrightarrow\:{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} } \\$$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{arcsin}\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}} {\int}}\frac{\mathrm{cos}^{\mathrm{2}} \:{u}}{\mathrm{2}+\mathrm{sin}^{\mathrm{2}} \:{u}}{du} \\$$$$\mathrm{third}\:\mathrm{step} \\$$$${v}=\mathrm{tan}\:{u}\:\Leftrightarrow\:{u}=\mathrm{arctan}\:{v}\:\Leftrightarrow\:{du}=\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{1}} \\$$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\left({v}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}\right)}{dv}= \\$$$$=\mathrm{6}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}}{dv}−\mathrm{2}\underset{\mathrm{0}} {\overset{\sqrt{\mathrm{2}}} {\int}}\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}{dv}= \\$$$$=\left[\sqrt{\mathrm{6}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{6}}{v}}{\mathrm{2}}\:−\mathrm{2arctan}\:{v}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} = \\$$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\pi−\mathrm{2arctan}\:\sqrt{\mathrm{2}} \\$$

Commented by bobhans last updated on 18/Aug/20

$$\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}=\:\mathrm{2cos}\:^{\mathrm{2}} {x}\: \\$$$${but}\:{the}\:{original}\:{equation}\:{in}\:{denumerator} \\$$$${is}\:\mathrm{cos}\:^{\mathrm{2}} {x}\: \\$$

Commented by Sarah85 last updated on 18/Aug/20

$$\mathrm{typo},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\$$