Question-143952

Question Number 143952 by akolade last updated on 19/Jun/21

Answered by Olaf_Thorendsen last updated on 20/Jun/21

C = ∫((coshx)/(coshx+sinhx)) dx S = ∫((sinhx)/(coshx+sinhx)) dx C+S = ∫dx = x+cst (1) C−S = ∫((coshx−sinhx)/(coshx+sinhx)) dx C−S = ∫(((coshx−sinhx)^2 )/(cosh^2 x−sinh^2 x)) dx C−S = ∫(((e^x +e^(−x) )/2)−((e^x −e^(−x) )/2))^2 dx C−S = ∫e^(−2x) dx = −(1/2)e^(−2x) +cst (2) (((1)+(2))/2) : C = −(1/4)e^(−2x) +(1/2)x+cst

$$\mathrm{C}\:=\:\int\frac{\mathrm{cosh}{x}}{\mathrm{cosh}{x}+\mathrm{sinh}{x}}\:{dx} \\ $$$$\mathrm{S}\:=\:\int\frac{\mathrm{sinh}{x}}{\mathrm{cosh}{x}+\mathrm{sinh}{x}}\:{dx} \\ $$$$\mathrm{C}+\mathrm{S}\:=\:\int{dx}\:=\:{x}+\mathrm{cst}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\int\frac{\mathrm{cosh}{x}−\mathrm{sinh}{x}}{\mathrm{cosh}{x}+\mathrm{sinh}{x}}\:{dx} \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\int\frac{\left(\mathrm{cosh}{x}−\mathrm{sinh}{x}\right)^{\mathrm{2}} }{\mathrm{cosh}^{\mathrm{2}} {x}−\mathrm{sinh}^{\mathrm{2}} {x}}\:{dx} \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\int\left(\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}−\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$$\mathrm{C}−\mathrm{S}\:=\:\int{e}^{−\mathrm{2}{x}} {dx}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} +\mathrm{cst}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\::\:\mathrm{C}\:=\:−\frac{\mathrm{1}}{\mathrm{4}}{e}^{−\mathrm{2}{x}} +\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{cst} \\ $$

Answered by Olaf_Thorendsen last updated on 20/Jun/21

C = ∫((coshx)/(coshx+sinhx)) dx C = ∫(((e^x +e^(−x) )/2)/(((e^x +e^(−x) )/2)+((e^x −e^(−x) )/2))) dx C = (1/2)∫((e^x +e^(−x) )/e^x ) dx C = (1/2)∫(1+e^(−2x) ) dx C = (1/2)(x−(1/2)e^(−2x) )+cst

$$\mathrm{C}\:=\:\int\frac{\mathrm{cosh}{x}}{\mathrm{cosh}{x}+\mathrm{sinh}{x}}\:{dx} \\ $$$$\mathrm{C}\:=\:\int\frac{\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}}{\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}+\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}}\:{dx} \\ $$$$\mathrm{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{e}^{{x}} +{e}^{−{x}} }{{e}^{{x}} }\:{dx} \\ $$$$\mathrm{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+{e}^{−\mathrm{2}{x}} \right)\:{dx} \\ $$$$\mathrm{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{x}} \right)+\mathrm{cst} \\ $$

Commented by akolade last updated on 20/Jun/21

$$\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{2xe}^{\mathrm{2x}} −\mathrm{1}}{\mathrm{4e}^{\mathrm{2x}} } \\ $$

Answered by mathmax by abdo last updated on 20/Jun/21

Φ=∫ ((ch(x))/(ch(x)+sh(x)))dx ⇒Φ=∫ (((e^x +e^(−x) )/2)/(((e^x +e^(−x) )/2)+((e^x −e^(−x) )/2)))dx =∫ ((e^x +e^(−x) )/(2e^x ))dx =(1/2)∫(1+e^(−2x) )dx =(x/2) +(1/2)∫ e^(−2x ) dx =(x/2)−(1/4)e^(−2x) +K

$$\Phi=\int\:\frac{\mathrm{ch}\left(\mathrm{x}\right)}{\mathrm{ch}\left(\mathrm{x}\right)+\mathrm{sh}\left(\mathrm{x}\right)}\mathrm{dx}\:\Rightarrow\Phi=\int\:\:\frac{\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}}{\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}+\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}}\mathrm{dx} \\ $$$$=\int\:\:\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2e}^{\mathrm{x}} }\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2x}} \right)\mathrm{dx}\:=\frac{\mathrm{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{e}^{−\mathrm{2x}\:} \mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{−\mathrm{2x}} \:+\mathrm{K} \\ $$

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