Integration Questions

Question Number 104609 by  M±th+et+s last updated on 22/Jul/20

$${prove}: \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{4}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{22}}{\mathrm{7}}−\pi \\$$

Answered by Dwaipayan Shikari last updated on 22/Jul/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{4}} \left(\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+\mathrm{4}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{4}{x}^{\mathrm{5}} +\frac{\mathrm{4}{x}^{\mathrm{6}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{4}} +{x}^{\mathrm{6}} −\mathrm{4}{x}^{\mathrm{5}} +\frac{\mathrm{4}{x}^{\mathrm{6}} +\mathrm{4}}{\mathrm{1}+{x}^{\mathrm{2}} }−\frac{\mathrm{4}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\left[\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}}−\frac{\mathrm{4}{x}^{\mathrm{6}} }{\mathrm{6}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\int\mathrm{4}\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)−\left[\mathrm{4}\:{tan}^{−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{4}−\mathrm{4}.\frac{\pi}{\mathrm{4}} \\$$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}−\mathrm{2}+\mathrm{4}−\pi \\$$$$=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{7}}−\pi \\$$$$=\frac{\mathrm{22}}{\mathrm{7}}−\pi\:\:\: \\$$$$\mathrm{It}\:\mathrm{proves}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\pi}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{smaller}}\:\boldsymbol{\mathrm{than}}\:\frac{\mathrm{22}}{\mathrm{7}} \\$$

Commented by Dwaipayan Shikari last updated on 22/Jul/20

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Commented by Ar Brandon last updated on 22/Jul/20

Great decomposition ��

Commented by  M±th+et+s last updated on 22/Jul/20

$${nice}\:{work}\:{sir} \\$$

Commented by Dwaipayan Shikari last updated on 22/Jul/20

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