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Question Number 51987 by maxmathsup by imad last updated on 01/Jan/19

$$calculate{\int }_0^{\frac{1}{2}}\sqrt{1-x^4}dx$$

Commented by Abdo msup. last updated on 02/Jan/19

$$letA={\int }_0^{\frac{1}{2}}\sqrt{1-x^4}dx\Rightarrow A={\int }_0^{\frac{1}{2}}\sqrt{1-x^2}\sqrt{1+x^2}dx$$$${=}_{x^2=cost}{\int }_1^{\frac{\pi }{3}}\sqrt{1-cost}\sqrt{1+cost}\frac{-sint}{2\sqrt{cost}}dt$$$$=-{\int }_1^{\frac{\pi }{3}}\sqrt{2}sin\left(\frac{t}{2}\right)\sqrt{2}cos\left(\frac{t}{2}\right)\frac{sint}{2\sqrt{cost}}dt$$$$=-{\int }_1^{\frac{\pi }{3}}sin\left(\frac{t}{2}\right)cos\left(\frac{t}{2}\right)\frac{sint}{\sqrt{cost}}dt$$$$=-{\int }_1^{\frac{\pi }{3}}\frac{1}{2}\frac{{sin}^{2}t}{\sqrt{cost}}dt=-\frac{1}{2}{\int }_1^{\frac{\pi }{3}}\frac{1-{cos}^{2}t}{\sqrt{cost}}dt$$$$=-\frac{1}{2}{\int }_1^{\frac{\pi }{3}}\frac{dt}{\sqrt{cost}}+{\int }_1^{\frac{\pi }{3}}\frac{{cos}^{2}t}{\sqrt{cost}}dt....$$$$becontinued...$$

Answered by peter frank last updated on 01/Jan/19

$$\int \sqrt{(1+x^2)(1-x^2)}$$$$x^2=\sin \theta \Rightarrow 2xdx=\cos \theta d\theta$$$$$$$$\int \sqrt{(1+\sin \theta )(1-\sin \theta )}dx$$$$\int \sqrt{1-\sin {}^2}\frac{\cos \theta d\theta }{2x}$$$$\frac{1}{2}\int \frac{\cos {}^2\theta d\theta }{x}$$$$\frac{1}{2}\int \frac{\cos {}^2\theta d\theta }{\sqrt{\sin \theta }}$$$$.....$$$$$$

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