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Question Number 59528 by Mr X pcx last updated on 11/May/19

$$letf\left(x\right)={\int }_0^1\frac{dt}{1+xch\left(t\right)}withxreal$$$$1)determineaexplicitformoff\left(x\right)$$$$2)findalsog\left(x\right)={\int }_0^1\frac{dt}{{(1+xch\left(t\right))}^2}$$$$3)calculate{\int }_0^1\frac{dt}{1+3ch\left(t\right)}and$$$${\int }_0^1\frac{dt}{{(1+3ch\left(t\right))}^2}$$

Commented by MJS last updated on 11/May/19

$$\mathrm{same}\mathrm{path}\mathrm{as}\mathrm{before}$$$$\int \frac{dt}{1+x\cosh t}=2\int \frac{{\mathrm{e}}^t}{x{\mathrm{e}}^{2t}+{2\mathrm{e}}^t+x}dt=\frac{2}{\sqrt{x^2-1}}\arctan \frac{x{\mathrm{e}}^t+1}{\sqrt{x^2-1}}+C$$$$\int \frac{dt}{{(1+x\cosh t)}^2}=4\int \frac{{\mathrm{e}}^{2t}}{{(x{\mathrm{e}}^{2t}+{2\mathrm{e}}^t+x)}^2}dt=-\frac{2}{{(x^2-1)}^{3/2}}\arctan \frac{x{\mathrm{e}}^t+1}{\sqrt{x^2-1}}-\frac{2({\mathrm{e}}^t+x)}{(x{\mathrm{e}}^{2t}+{2\mathrm{e}}^t+x)(x^2-1)}$$

Commented by maxmathsup by imad last updated on 11/May/19

$$thankssir$$

Commented by maxmathsup by imad last updated on 12/May/19

$$1)wehavef\left(x\right)={\int }_0^1\frac{dt}{1+x\frac{e^t+e^{-t}}{2}}={\int }_0^1\frac{2dt}{2+{xe}^t+x{e}^{-t}}$$$${=}_{e^t=u}{\int }_1^e\frac{2}{2+xu+{xu}^{-1}}\frac{du}{u}={\int }_1^e\frac{2du}{2u+{xu}^2+x}$$$$={\int }_1^e\frac{2du}{{xu}^2+2u+x}polesofF\left(u\right)=\frac{2}{{xu}^2+2u+x}$$$${xu}^2+2u+x=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1-x^{2}case11-x^2>0\Rightarrow \mid x\mid <1\Rightarrow u_1=\frac{-1+\sqrt{1-x^2}}{x}$$$$u_2=\frac{-1-\sqrt{1-x^2}}{x}(wesupposex\ne 0)\Rightarrow F\left(u\right)=\frac{2}{x(u-u_1)(u-u_2)}$$$$=\frac{2}{x(u_1-u_2)}(\frac{1}{u-u_1}-\frac{1}{u-u_2})=\frac{2}{x\frac{2\sqrt{1-x^2}}{x}}(\frac{1}{u-u_1}-\frac{1}{u-u_2})=\frac{1}{\sqrt{1-x^2}}(\frac{1}{u-u_1}-\frac{1}{u-u_2})$$$$f\left(x\right)={\int }_1^{e}F\left(u\right)du=\frac{1}{\sqrt{1-x^2}}{[ln\mid \frac{u-u_1}{u-u_2}\mid ]}_1^e=\frac{1}{\sqrt{1-x^2}}\{ln\mid \frac{e-u_1}{e-u_2}\mid -ln\mid \frac{1-u_1}{1-u_2}\mid \}$$$$=\frac{1}{\sqrt{1-x^2}}\{ln\mid \frac{e-\frac{-1+\sqrt{1-x^2}}{x}}{e-\frac{-1-\sqrt{1-x^2}}{x}}\mid -ln\mid \frac{1-\frac{-1+\sqrt{1-x^2}}{x}}{1-\frac{-1-\sqrt{1-x^2}}{x}}\mid \}$$$$=\frac{1}{\sqrt{1-x^2}}\{ln\mid \frac{ex+1-\sqrt{1-x^2}}{ex+1+\sqrt{1-x^2}}\mid -ln\mid \frac{x+1-\sqrt{1-x^2}}{x+1+\sqrt{1-x^2}}\mid \}$$

Commented by maxmathsup by imad last updated on 12/May/19

$$case21-x^2<0\Rightarrow x^2-1>0\Rightarrow F\left(u\right)=\frac{2}{{xu}^2+2u+x}=\frac{2}{x(u^2+2\frac{u}{x}+1)}$$$$=\frac{2}{x\{u^2+2\frac{u}{x}+\frac{1}{x^2}+1-\frac{1}{x^2})}=\frac{2}{x\{{(u+\frac{1}{x})}^2+\frac{x^2-1}{x^2}\}}$$$$=\frac{2}{x\{{(u+\frac{1}{x})}^2+{\left(\frac{\sqrt{x^2-1}}{\mid x\mid }\right)}^2}\Rightarrow chang.u+\frac{1}{x}=\frac{\sqrt{x^2-1}}{\mid x\mid }\alpha give\alpha =\frac{xu+1}{x}\frac{\mid x\mid }{\sqrt{x^2-1}}$$$$f\left(x\right)={\int }_1^{e}F\left(u\right)du=\frac{2}{x}{\int }_{\xi \left(x\right)\frac{x+1}{\sqrt{x^2-1}}}^{\xi \left(x\right)\frac{ex+1}{\sqrt{x^2-1}}}\frac{1}{\frac{x^2-1}{\mid x\mid }(1+{\alpha }^2)}\frac{\sqrt{x^2-1}}{\mid x\mid }d\alpha$$$$=\frac{2}{x\sqrt{x^2-1}}{\left[arctan\left(\alpha \right)\right]}_{\frac{x+1}{\sqrt{x^2-1}}\xi \left(x\right)}^{\frac{ex+1}{\sqrt{x^2-1}}\xi \left(x\right)}$$$$=\frac{2}{x\sqrt{x^2-1}}\{arctan\left(\frac{ex+1}{\sqrt{x^2-1}}\xi \left(x\right)\right)-arctan\left(\frac{x+1}{\sqrt{x^2-1}}\xi \left(x\right)\right)\}with$$$$\xi \left(x\right)=1ifx>0and\xi \left(x\right)=-1ifx<0.$$

Commented by maxmathsup by imad last updated on 12/May/19

$$3){\int }_0^1\frac{dt}{1+3ch\left(t\right)}=f\left(3\right)\left(case2\right)$$$$=\frac{2}{3\sqrt{8}}\{arctan\left(\frac{3e+1}{\sqrt{8}}\right)-arctan\left(\frac{4}{\sqrt{8}}\right)\}$$$$=\frac{1}{3\sqrt{2}}\{arctan\left(\frac{3e+1}{2\sqrt{2}}\right)-arctan\left(\sqrt{2}\right)\}.$$

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