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Question Number 29972 by abdo imad last updated on 14/Feb/18

$$letgive\mid x\mid <1find{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-x^2{cos}^2\theta }}.$$

Commented by abdo imad last updated on 18/Feb/18

$$wehave{(1+u^{})}^{\alpha }=1+\alpha u+\frac{\alpha (\alpha -1)}{2!}{u}^2+...$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{\alpha (\alpha -1)...(\alpha -n+1)}{n!}{u}^{n}with\mid u\mid <1wehave$$$$\mid xcos\theta \mid <1\Rightarrow \frac{1}{\sqrt{1-x^2{cos}^2\theta }}={(1-{\left(xcos\theta \right)}^2)}^{-\frac{1}{2}}$$$$=1+\sum _{n=1}^{\mathrm{\infty }}\frac{(-\frac{1}{2})(-\frac{1}{2}-1)...(-\frac{1}{2}-n+1)}{n!}{x}^{2n}{cos}^{2n}\theta let$$$$put{A}_n=(-\frac{1}{2})(-\frac{1}{2}-1)...(-\frac{1}{2}-n+1)\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-x^2{cos}^2\theta }}=\frac{\pi }{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{A_n}{n!}{x}^{2n}{\int }_0^{\frac{\pi }{2}}{cos}^{2n}\theta d\theta but$$$$w_n={\int }_0^{\frac{\pi }{2}}{cos}^{2n}\theta d\theta isknown\left(wallisintegral\right)so$$$${\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-x^2{cos}^2\theta }}=\frac{\pi }{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{A_n{B}_n}{n!}{x}^{2n}.$$

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