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Question Number 40151 by maxmathsup by imad last updated on 16/Jul/18

$$letF\left(x\right)={\int }_0^{\frac{\pi }{2}}cos\left(xsint\right)dt$$$$1)provethat\mathrm{\forall }u\in R1-\frac{u^2}{2}⩽cosu⩽1-\frac{u^2}{2}+\frac{u^4}{24}$$$$2)provethat\frac{\pi }{2}(1-\frac{x^2}{4})⩽F\left(x\right)⩽\frac{\pi }{2}(1-\frac{x^2}{4}+\frac{x^4}{64})$$

Commented by math khazana by abdo last updated on 19/Jul/18

$$1)wehavecos\left(u\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{u}^{2n}}{\left(2n\right)!}$$$$=1-\frac{u^2}{2}+\frac{u^4}{\left(4\right)!}-\frac{u^6}{6!}+....\mathrm{\forall }u\in R\Rightarrow$$$$1-\frac{u^2}{2}⩽cosu⩽1-\frac{u^2}{2}+\frac{u^4}{24}$$$$2)weget1-\frac{{\left(xsint\right)}^2}{2}⩽cos\left(xsint\right)⩽1-\frac{{\left(xsint\right)}^2}{2}$$$$+\frac{{\left(xsint\right)}^4}{24}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}(1-\frac{x^2}{2}{sin}^{2}t)dt⩽{\int }_0^{\frac{\pi }{2}}cos\left(xsint\right)dt$$$$⩽{\int }_0^{\frac{\pi }{2}}(1-\frac{x^2{sin}^{2}t}{2}+\frac{x^4{sin}^4}{24})dtbut$$$${\int }_0^{\frac{\pi }{2}}(1-\frac{x^2}{2}{sin}^{2}t)dt=\frac{\pi }{2}-\frac{x^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{1-cos\left(2t\right)}{2}dt$$$$=\frac{\pi }{2}-\frac{x^2}{4}\frac{\pi }{2}=\frac{\pi }{2}(1-\frac{x^2}{4})also$$$${\int }_0^{\frac{\pi }{2}}(1-\frac{x^2{sin}^{2}t}{2}+\frac{x^4{sin}^{4}t}{24})dt$$$$=\frac{\pi }{2}(1-\frac{x^2}{4})+\frac{x^4}{24}{\int }_0^{\frac{\pi }{2}}{\left(\frac{1-cos\left(2t\right)}{2}\right)}^{2}dt$$$$=\frac{\pi }{2}(1-\frac{x^2}{4})+\frac{x^4}{96}{\int }_0^{\frac{\pi }{2}}(1-2cos\left(2t\right)+{cos}^2\left(2t\right))dt$$$$=\frac{\pi }{2}(1-\frac{x^2}{2})+\frac{x^4}{96}\frac{\pi }{2}+\frac{x^4}{96}{\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(4t\right)}{2}dt$$$$=\frac{\pi }{2}(1-\frac{x^2}{2})+\frac{\pi }{2}(\frac{1}{96}+\frac{1}{2.96})x^4$$$$=\frac{\pi }{2}(1-\frac{x^2}{2})+\frac{x^4}{64}.\frac{\pi }{2}=\frac{\pi }{2}(1-\frac{x^2}{2}+\frac{x^4}{64})\Rightarrow$$$$tberesultisproved.$$$$$$$$$$$$$$$$$$$$$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18

$$p+iq={\int }_0^{\frac{\mathrm{\Pi }}{2}}{e}^{ixsint}dt$$$${\int }_0^{\frac{\mathrm{\Pi }}{2}}1+ixsint+\frac{i^2{x}^2{sin}^{2}t}{2!}+\frac{i^3{x}^3{sin}^{3}t}{3!}+\frac{i^4{x}^4{sin}^{4}t}{4!}+..$$$$p=F\left(x\right)=$$$${\int }_0^{\frac{\mathrm{\Pi }}{2}}1-\frac{x^2{sin}^{2}t}{4}+\frac{x^4{sin}^{4}t}{24}+..dt$$$$1⩾{sin}^{2}t⩾0...1⩾{sin}^{2k}t⩾02k=even$$$$so{\int }_0^{\frac{\mathrm{\Pi }}{2}}1-\frac{x^2}{4}dt⩽F\left(x\right)⩽{\int }_0^{\frac{\mathrm{\Pi }}{2}}1-\frac{x^2}{4}+\frac{x^4}{24}dt$$$$(1-\frac{x^2}{4})\frac{\mathrm{\Pi }}{2}⩽F\left(x\right)⩽(1-\frac{x^2}{4}+\frac{x^4}{24})\frac{\mathrm{\Pi }}{2}$$$$$$

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