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Question Number 65924 by mathmax by abdo last updated on 05/Aug/19

$$calculate{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{x}{sinx}dx$$

Commented by mathmax by abdo last updated on 07/Aug/19

$$letI={\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{x}{sinx}dxletfindapproximatevaluewehave$$$$I=2{\int }_0^{\frac{\pi }{6}}\frac{x}{sinx}dxbutsinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}withradiusR=+\mathrm{\infty }$$$$\Rightarrow sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots .\Rightarrow x-\frac{x^3}{3!}⩽sinx⩽x\Rightarrow \frac{1}{x}⩽\frac{1}{sinx}⩽\frac{1}{x-\frac{x^3}{6}}$$$$\Rightarrow 1⩽\frac{x}{sinx}⩽\frac{1}{1-\frac{x^2}{6}}forx\in ]0,\frac{\pi }{6}]\Rightarrow {\int }_0^{\frac{\pi }{6}}1dx⩽{\int }_0^{\frac{\pi }{6}}\frac{x}{sinx}dx⩽{\int }_0^{\frac{\pi }{6}}\frac{6dx}{6-x^2}$$$$\frac{\pi }{3}⩽2{\int }_0^{\frac{\pi }{6}}\frac{x}{sinx}dx⩽12{\int }_0^{\frac{\pi }{6}}\frac{dx}{6-x^2}\Rightarrow \frac{\pi }{3}⩽I⩽12{\int }_0^{\frac{\pi }{6}}\frac{dx}{6-x^2}$$$${\int }_0^{\frac{\pi }{6}}\frac{dx}{6-x^2}=-{\int }_0^{\frac{\pi }{6}}\frac{dx}{(x-\sqrt{6})(x+\sqrt{6})}=-\frac{1}{2\sqrt{6}}{\int }_0^{\frac{\pi }{6}}\{\frac{1}{x-\sqrt{6}}-\frac{1}{x+\sqrt{6}}\}dx$$$$=-\frac{1}{2\sqrt{6}}{[ln\mid \frac{x-\sqrt{6}}{x+\sqrt{6}}\mid ]}_0^{\frac{\pi }{6}}=-\frac{1}{2\sqrt{6}}ln\mid \frac{\frac{\pi }{6}-\sqrt{6}}{\frac{\pi }{6}+\sqrt{6}}\mid$$$$=-\frac{1}{2\sqrt{6}}ln\mid \frac{\pi -6\sqrt{6}}{\pi +6\sqrt{6}}\mid =\frac{1}{2\sqrt{6}}ln\left(\frac{\pi +6\sqrt{6}}{6\sqrt{6}-\pi }\right)\Rightarrow \frac{\pi }{3}⩽I⩽\sqrt{6}ln\left(\frac{6\sqrt{6}+\pi }{6\sqrt{6}-\pi }\right)$$$$let{v}_0=\frac{\pi }{6}+\frac{\sqrt{6}}{2}ln\left(\frac{6\sqrt{6}+\pi }{6\sqrt{6}-\pi }\right)$$$$v_{0}isabetterapproximationforI.$$$$$$

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