Question Number 176540 by yaslm last updated on 20/Sep/22
Answered by Peace last updated on 21/Sep/22
$$\mathcal{F}\left({x}\right)=\int\frac{{dx}}{{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}={tg}\left({x}\right)\frac{\mathrm{1}}{\:\sqrt{{cos}\left({x}\right)}}−\int{tg}\left({x}\right).\frac{{sin}\left({x}\right)}{\mathrm{2}\sqrt{{cos}\left({x}\right)}{cos}\left({x}\right)} \\ $$$$−\int\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)\sqrt{{cos}\left({x}\right)}}{dx} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\mathcal{F}\left({x}\right)=\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)\sqrt{{cos}\left({x}\right)}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{cos}\left({x}\right)}} \\ $$$$\int\frac{{dx}}{\:\sqrt{{cos}\left({x}\right)}}=\int\frac{{dx}}{\:\sqrt{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}}}=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}} \\ $$$$=\int\frac{\mathrm{2}{dy}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({y}\right)}}=\mathrm{2}{F}\left({y}\mid\mathrm{2}\right)=\mathrm{2}{F}\left(\frac{{x}}{\mathrm{2}}\mid\mathrm{2}\right) \\ $$$${F}\left({x}\mid{a}^{\mathrm{2}} \right)=\int_{\mathrm{0}} ^{{x}} \frac{{ds}}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({s}\right)}}\:\:{Eleptic}\:{integral} \\ $$$$\mathcal{F}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}}\left({tg}\left({x}\right){sec}\left({x}\right)+{F}\left(\frac{{x}}{\mathrm{2}}\mid\mathrm{2}\right)\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 22/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Peace last updated on 23/Sep/22
$${withe}\:{Pleasur} \\ $$