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Question Number 176679 by mnjuly1970 last updated on 24/Sep/22

$$$$$$Eeasyintegral....$$$$\mathbf{\Omega }={\int }_{-{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx}^{{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx}{sin}^2\left(t\right).{ln}^3(t+\sqrt{1+t^2})dt$$$$---m.n---$$

Answered by Peace last updated on 24/Sep/22

$$ln(-t+\sqrt{1+t^2})=-ln(t+\sqrt{1+t^2})$$$$\mathrm{\Omega }=0$$

Answered by mahdipoor last updated on 25/Sep/22

$$-f\left(a\right)=-[{sin}^2\left(a\right).{ln}^3(a+\sqrt{a^2+1})]=$$$${sin}^2\left(a\right).{[-ln(a+\sqrt{a^2+1})]}^3=$$$${sin}^2\left(a\right).{ln}^3\left(\frac{1}{a+\sqrt{a^2+1}}\right)=$$$${sin}^2\left(a\right).{ln}^3\left(\frac{-a+\sqrt{a^2+1}}{[a+\sqrt{a^2+1}][-a+\sqrt{a^2+1}]}\right)=$$$${sin}^2(-a).{ln}^3(-a+\sqrt{{(-a)}^2+1})=f(-a)$$$$\Rightarrow \Rightarrow$$$$\mathrm{\forall }a,f\left(\xi \right).\delta x+f(-\xi ).\delta x=0$$$$\Rightarrow \Rightarrow$$$$get{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=m$$$${\xi }_i=i.\delta x$$$$\Rightarrow \Rightarrow$$$$0=\underset{\delta x\to 0}{\lim }\underset{i=0}{\sum ^{m/\delta x}}f(-{\xi }_i).\delta x+f\left({\xi }_i\right).\delta x=$$$$\underset{\delta x\to 0}{\lim }\underset{-m/\delta x}{\sum ^{m/\delta x}}f\left({\xi }_i\right).\delta x={\int }_{-m}^{m}f\left(x\right)dx$$$$$$

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