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Question Number 66062 by mathmax by abdo last updated on 08/Aug/19

$$letf\left(x\right)={\int }_0^1\frac{dt}{ch\left(t\right)+xsh\left(t\right)}$$$$1)findaexplicitformoff\left(x\right)$$$$2)determineg\left(x\right)={\int }_0^1\frac{dt}{{(ch\left(t\right)+xsh\left(t\right))}^2}$$$$3)calculate{\int }_0^1\frac{dt}{ch\left(t\right)+3sh\left(t\right)}and{\int }_0^1\frac{dt}{{\{ch\left(t\right)+3sh\left(t\right)\}}^2}$$

Commented by mathmax by abdo last updated on 11/Aug/19

$$1)wehavef\left(x\right)={\int }_0^1\frac{dt}{ch\left(t\right)+xsh\left(t\right)}\Rightarrow$$$$f\left(x\right)={\int }_0^1\frac{2dt}{e^t+e^{-t}+x(e^t-e^{-t})}={\int }_0^1\frac{2dt}{(1+x)e^t+(1-x)e^{-t}}$$$$changement{e}^t=ugivef\left(x\right)={\int }_1^e\frac{2}{(1+x)u+(1-x)u^{-1}}\frac{du}{u}$$$$={\int }_1^e\frac{2du}{(1+x)u^2+(1-x)}=\frac{2}{1+x}{\int }_1^e\frac{du}{u^2+\frac{1-x}{1+x}}$$$$case1\frac{1-x}{1+x}>0\Rightarrow \mid x\mid <1wedothechangementu=\sqrt{\frac{1-x}{1+x}}z\Rightarrow$$$$f\left(x\right)=\frac{2}{1+x}{\int }_{\sqrt{\frac{1+x}{1-x}}}^{e\sqrt{\frac{1+x}{1-x}}}\frac{1+x}{1-x}\frac{1}{1+z^2}\sqrt{\frac{1-x}{1+x}}dz$$$$=\frac{2}{1-x}\frac{\sqrt{1-x}}{\sqrt{1+x}}{\left[arctan\left(z\right)\right]}_{\sqrt{\frac{1+x}{1-x}}}^{e\sqrt{\frac{1+x}{1-x}}}$$$$=\frac{2}{\sqrt{1-x^2}}\{arctan\left(e\sqrt{\frac{1+x}{1-x}}\right)-arctan\left(\sqrt{\frac{1+x}{1-x}}\right)\}$$$$case2\frac{1-x}{1-x}<0\Rightarrow \mid x\mid >1\Rightarrow f\left(x\right)=\frac{2}{x+1}{\int }_1^e\frac{du}{u^2-{\left(\sqrt{\frac{x-1}{x+1}}\right)}^2}$$$${=}_{u=\sqrt{\frac{x-1}{x+1}}z}\frac{2}{x+1}{\int }_{\sqrt{\frac{x+1}{x-1}}}^{e\sqrt{\frac{x+1}{x-1}}}\frac{x+1}{x-1}\frac{1}{z^2-1}\sqrt{\frac{x-1}{x+1}}dz$$$$=\frac{1}{x-1}\frac{\sqrt{x-1}}{\sqrt{x+1}}{\int }_{\sqrt{\frac{x+1}{x-1}}}^{e\sqrt{\frac{x+1}{x-1}}}\{\frac{1}{z-1}-\frac{1}{z+1}\}dz$$$$=\frac{1}{\sqrt{x^2-1}}{[ln\mid \frac{z-1}{z+1}\mid ]}_{\sqrt{\frac{x+1}{x-1}}}^{e\sqrt{\frac{x+1}{x-1}}}=\frac{1}{\sqrt{x^2-1}}\{ln\mid \frac{\frac{e\sqrt{x+1}}{\sqrt{x-1}}-1}{\frac{e\sqrt{x+1}}{\sqrt{x-1}}+1}\mid$$$$-ln\mid \frac{\frac{\sqrt{x+1}}{\sqrt{x-1}}-1}{\frac{\sqrt{x+1}}{\sqrt{x-1}}+1}\mid \}\Rightarrow$$$$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}\{ln\mid \frac{e\sqrt{x+1}-\sqrt{x-1}}{e\sqrt{x+1}+\sqrt{x-1}}\mid -ln\mid \frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\mid \}$$

Commented by mathmax by abdo last updated on 11/Aug/19

$$2)sorryg\left(x\right)={\int }_0^1\frac{sh\left(t\right)dt}{{(ch\left(t\right)+xsh\left(t\right))}^2}$$$$wehave{f}^{{}^{\prime }}\left(x\right)=-{\int }_1^1\frac{sh\left(t\right)}{{(ch\left(t\right)+xsh\left(t\right))}^2}=-g\left(x\right)\Rightarrow$$$$g\left(x\right)=-f^{{}^{\prime }}\left(x\right)resttocalculate{f}^{{}^{\prime }}\left(x\right)$$$$$$

Commented by mathmax by abdo last updated on 11/Aug/19

$$3){\int }_0^1\frac{dt}{ch\left(t\right)+3sh\left(t\right)}=f\left(3\right)=\frac{1}{2\sqrt{2}}\{ln\left(\frac{2e-\sqrt{2}}{2e+\sqrt{2}}\right)-ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)\}$$

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