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# calculate-pi-6-pi-6-x-sinx-dx-

Question Number 65924 by mathmax by abdo last updated on 05/Aug/19
$${calculate}\:\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{x}}{{sinx}}{dx}\: \\$$
Commented by mathmax by abdo last updated on 07/Aug/19
$${let}\:{I}\:=\int_{−\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{x}}{{sinx}}{dx}\:\:{let}\:{find}\:{approximate}\:{value}\:{we}\:{have} \\$$$${I}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{x}}{{sinx}}{dx}\:\:\:\:\:{but}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{with}\:{radiusR}=+\infty \\$$$$\Rightarrow{sinx}\:={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−….\Rightarrow{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\leqslant{sinx}\:\leqslant{x}\:\Rightarrow\frac{\mathrm{1}}{{x}}\leqslant\frac{\mathrm{1}}{{sinx}}\leqslant\frac{\mathrm{1}}{{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}} \\$$$$\left.\Rightarrow\left.\mathrm{1}\leqslant\frac{{x}}{{sinx}}\leqslant\frac{\mathrm{1}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}\:\:\:{for}\:{x}\in\right]\mathrm{0},\frac{\pi}{\mathrm{6}}\right]\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{1}{dx}\:\leqslant\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{x}}{{sinx}}{dx}\leqslant\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{6}{dx}}{\mathrm{6}−{x}^{\mathrm{2}} } \\$$$$\frac{\pi}{\mathrm{3}}\leqslant\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\frac{{x}}{{sinx}}{dx}\:\leqslant\mathrm{12}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{dx}}{\mathrm{6}−{x}^{\mathrm{2}} }\:\Rightarrow\frac{\pi}{\mathrm{3}}\leqslant{I}\:\leqslant\mathrm{12}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{dx}}{\mathrm{6}−{x}^{\mathrm{2}} } \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{dx}}{\mathrm{6}−{x}^{\mathrm{2}} }\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\frac{{dx}}{\left({x}−\sqrt{\mathrm{6}}\right)\left({x}+\sqrt{\mathrm{6}}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\left\{\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{6}}}−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{6}}}\right\}{dx} \\$$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}\left[{ln}\mid\frac{{x}−\sqrt{\mathrm{6}}}{{x}+\sqrt{\mathrm{6}}}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}{ln}\mid\frac{\frac{\pi}{\mathrm{6}}−\sqrt{\mathrm{6}}}{\frac{\pi}{\mathrm{6}}+\sqrt{\mathrm{6}}}\mid \\$$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}{ln}\mid\frac{\pi−\mathrm{6}\sqrt{\mathrm{6}}}{\pi+\mathrm{6}\sqrt{\mathrm{6}}}\mid\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}{ln}\left(\frac{\pi+\mathrm{6}\sqrt{\mathrm{6}}}{\mathrm{6}\sqrt{\mathrm{6}}−\pi}\right)\:\Rightarrow\frac{\pi}{\mathrm{3}}\leqslant\:{I}\:\leqslant\sqrt{\mathrm{6}}{ln}\left(\frac{\mathrm{6}\sqrt{\mathrm{6}}+\pi}{\mathrm{6}\sqrt{\mathrm{6}}−\pi}\right) \\$$$${let}\:{v}_{\mathrm{0}} =\frac{\pi}{\mathrm{6}}\:+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{ln}\left(\frac{\mathrm{6}\sqrt{\mathrm{6}}+\pi}{\mathrm{6}\sqrt{\mathrm{6}}−\pi}\right) \\$$$${v}_{\mathrm{0}} {is}\:{a}\:{better}\:{approximation}\:{for}\:{I}\:. \\$$$$\\$$