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

$$prove:$$$${\int }_0^1\frac{x^4{(1-x)}^4}{1+x^2}dx=\frac{22}{7}-\pi$$

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

$${\int }_0^1\frac{x^4{(1-2x+x^2)}^2}{1+x^2}dx$$$${\int }_0^1\frac{x^4({(1+x^2)}^2-4x(1+x^2)+4x^2)}{1+x^2}dx$$$${\int }_0^1{x}^4(1+x^2)-4x^5+\frac{4x^6}{1+x^2}dx$$$${\int }_0^1{x}^4+x^6-4x^5+\frac{4x^6+4}{1+x^2}-\frac{4}{1+x^2}dx$$$${[\frac{x^5}{5}+\frac{x^7}{7}-\frac{4x^6}{6}]}_0^1+\int 4(x^4-x^2+1)-{\left[4{tan}^{-1}x\right]}_0^1$$$$\frac{1}{5}+\frac{1}{7}-\frac{2}{3}+\frac{4}{5}-\frac{4}{3}+4-4.\frac{\pi }{4}$$$$=1+\frac{1}{7}-2+4-\pi$$$$=3+\frac{1}{7}-\pi$$$$=\frac{22}{7}-\pi$$$$\mathrm{It}\mathrm{proves}\mathbf{that}\mathit{\pi }\mathbf{is}\mathbf{smaller}\mathbf{than}\frac{22}{7}$$

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

$$niceworksir$$

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

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