Question Number 216042 by ajfour last updated on 26/Jan/25
$$\left({i}\right)\:\:\:\int\mathrm{sec}\:^{\mathrm{5}} \theta{d}\theta \\ $$$$\left({ii}\right)\:\:\int\:\frac{\sqrt{\mathrm{tan}\:\theta}\:{d}\theta}{\mathrm{cos}\:\theta} \\ $$
Commented by ajfour last updated on 26/Jan/25
https://youtu.be/VqEdd_VGxGI?si=VnVRn7bW5wENyi9Y
Find radius of circle inscribed in half ellipse.
Commented by Ghisom last updated on 26/Jan/25
$$\int\frac{\sqrt{\mathrm{tan}\:\theta}}{\mathrm{cos}\:\theta}{d}\theta\:\mathrm{leads}\:\mathrm{to}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{including} \\ $$$$\mathrm{the}\:\mathrm{Hypergeometrical}\:\mathrm{Function}\:_{\mathrm{2}} {F}_{\mathrm{1}} \\ $$
Answered by Ghisom last updated on 26/Jan/25
![∫sec^5 θ dθ= [t=sin θ] =−∫(dt/((t^2 −1)^3 ))= [Ostrogradski′s Method] =−((t(3t^2 −5))/(8(t^2 −1)^2 ))−(3/8)∫(dt/(t^2 −1))= =(3/(16))ln ∣((t+1)/(t−1))∣ −((t(3t^2 −5))/(8(t^2 −1)^2 ))=...](https://www.tinkutara.com/question/Q216044.png)
$$\int\mathrm{sec}^{\mathrm{5}} \:\theta\:{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sin}\:\theta\right] \\ $$$$=−\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=−\frac{{t}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{8}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\mathrm{ln}\:\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\mid\:−\frac{{t}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{8}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=… \\ $$
Commented by ajfour last updated on 26/Jan/25
$${okay},\:{thanks}. \\ $$
Answered by MathematicalUser2357 last updated on 27/Jan/25
$$\left({i}\right)\:\frac{\mathrm{1}}{\mathrm{16}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{11}\:\mathrm{sin}\:\theta+\mathrm{3}\:\mathrm{sin}\:\mathrm{3}\theta\right)\:\mathrm{sec}^{\mathrm{4}} \theta−\mathrm{6}\:\mathrm{log}\left(\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)+\mathrm{6}\:\mathrm{log}\left(\mathrm{sin}\:\frac{\theta}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\right\}+{C} \\ $$$$\left({ii}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{cos}\:\theta\:\mathrm{tan}^{\mathrm{3}/\mathrm{2}} \theta\sqrt{\mathrm{sec}^{\mathrm{2}} \theta}\: \\ $$
Commented by MathematicalUser2357 last updated on 27/Jan/25
Where 삼각사.초기하함수2F1(...) Represents Hypergeometric function 2F1