Question Number 21680 by Arnab Maiti last updated on 30/Sep/17
$$\int\sqrt{\mathrm{sec}\theta}\:\mathrm{d}\theta \\ $$
Answered by alex041103 last updated on 30/Sep/17
$${First}: \\ $$$$\sqrt{{sec}\theta}=\sqrt{\frac{\mathrm{1}}{{cos}\theta}}=\frac{\mathrm{1}}{\:\sqrt{{cos}\theta}} \\ $$$${We}\:{know}\:{that}\:{cos}\left(\mathrm{2}\theta\right)=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\Rightarrow\int\sqrt{{sec}\theta}{d}\theta=\int\frac{{d}\theta}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}} \\ $$$${We}\:{make}\:{the}\:{substitution} \\ $$$$\varphi=\theta/\mathrm{2}\:\left(\mathrm{2}\varphi=\theta\right)\:{and}\:{d}\theta=\mathrm{2}{d}\varphi \\ $$$$\Rightarrow\int\sqrt{{sec}\theta}{d}\theta=\int\frac{\mathrm{2}{d}\varphi}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \varphi}} \\ $$$${But}\:{integrals}\:{of}\:{the}\:{kind} \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{ksin}^{\mathrm{2}} {x}}}\:{are}\:{well}\:{known}\:{to}\:{be}\: \\ $$$${elliptic}\:{integral}\:{of}\:{first}\:{kind} \\ $$$${and}\:\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{ksin}^{\mathrm{2}} {x}}}\:=\:{F}\left({x}\mid{k}\right) \\ $$$$\Rightarrow\int\sqrt{{sec}\theta}{d}\theta=\mathrm{2}{F}\left(\varphi\mid\mathrm{2}\right)=\mathrm{2}{F}\left(\theta/\mathrm{2}\mid\mathrm{2}\right)+{C} \\ $$$${Ans}.\:\int\sqrt{{sec}\theta}\:{d}\theta\:=\:\mathrm{2}{F}\left(\theta/\mathrm{2}\mid\mathrm{2}\right)+{C} \\ $$