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Question Number 176540 by yaslm last updated on 20/Sep/22

Answered by Peace last updated on 21/Sep/22

$$\mathcal{F}\left(x\right)=\int \frac{dx}{{cos}^2\left(x\right)\sqrt{cos\left(x\right)}}=tg\left(x\right)\frac{1}{\sqrt{cos\left(x\right)}}-\int tg\left(x\right).\frac{sin\left(x\right)}{2\sqrt{cos\left(x\right)}cos\left(x\right)}$$$$-\int \frac{{sin}^2\left(x\right)}{2{cos}^2\left(x\right)\sqrt{cos\left(x\right)}}dx$$$$\frac{3}{2}\mathcal{F}\left(x\right)=\frac{sin\left(x\right)}{cos\left(x\right)\sqrt{cos\left(x\right)}}+\frac{1}{2}\int \frac{dx}{\sqrt{cos\left(x\right)}}$$$$\int \frac{dx}{\sqrt{cos\left(x\right)}}=\int \frac{dx}{\sqrt{2{cos}^2\left(\frac{x}{2}\right)-1}}=\int \frac{dx}{\sqrt{1-2{sin}^2\left(\frac{x}{2}\right)}}$$$$=\int \frac{2dy}{\sqrt{1-2{sin}^2\left(y\right)}}=2F(y\mid 2)=2F(\frac{x}{2}\mid 2)$$$$F(x\mid a^2)={\int }_0^x\frac{ds}{\sqrt{1-a^2{sin}^2\left(s\right)}}Elepticintegral$$$$\mathcal{F}\left(x\right)=\frac{2}{3}(tg\left(x\right)sec\left(x\right)+F(\frac{x}{2}\mid 2))+c$$$$$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 22/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by Peace last updated on 23/Sep/22

$$withePleasur$$

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