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Question Number 216042 by ajfour last updated on 26/Jan/25

$$\left(i\right)\int \sec {}^5\theta d\theta$$$$\left(ii\right)\int \frac{\sqrt{\tan \theta }d\theta }{\cos \theta }$$

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

$$\int \frac{\sqrt{\tan \theta }}{\cos \theta }d\theta \mathrm{leads}\mathrm{to}\mathrm{a}\mathrm{solution}\mathrm{including}$$$$\mathrm{the}\mathrm{Hypergeometrical}\mathrm{Function}{}_2{F}_1$$

Answered by Ghisom last updated on 26/Jan/25

$$\int {\sec }^5\theta d\theta =$$$$[t=\sin \theta ]$$$$=-\int \frac{dt}{{(t^2-1)}^3}=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=-\frac{t(3t^2-5)}{8{(t^2-1)}^2}-\frac{3}{8}\int \frac{dt}{t^2-1}=$$$$=\frac{3}{16}\ln \mid \frac{t+1}{t-1}\mid -\frac{t(3t^2-5)}{8{(t^2-1)}^2}=...$$

Commented by ajfour last updated on 26/Jan/25

$$okay,thanks.$$

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