Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

OR ((2xe^(2x) −1)/(4e^(2x) ))

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

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com