Evaluate: \(\int\cfrac{dx}{(e^x+e^{-x})}\)∫ dx/(ex + e-x)

1.	Evaluate: \(\int\cfrac{dx}{(e^x+e^{-x})}\)∫ dx/(ex + e-x)
Answer» To find: \(\int\cfrac{dx}{(e^x+e^{-x})}\) Formula Used:\(\int\cfrac{dx}{1+x^2}=tan^{-1}x\) Given equation is: \(\int\cfrac{dx}{(e^x+e^{-x})}=\int\cfrac{e^x dx}{(e^2x+1)}\)....(1) Let y = e^x … (1) Differentiating both sides, we get dy = e^x dx Substituting in (1), \(\int\cfrac{dy}{y^2+1}\) ⇒ tan^-1 y From (1), ⇒ tan^-1 (e^x) Therefore, \(\int\cfrac{dx}{(e^x+e^{-x})}=tan^{-1}(e^x)+c\)

Answer»

To find: \(\int\cfrac{dx}{(e^x+e^{-x})}\)

Formula Used:\(\int\cfrac{dx}{1+x^2}=tan^{-1}x\)

Given equation is:

\(\int\cfrac{dx}{(e^x+e^{-x})}=\int\cfrac{e^x dx}{(e^2x+1)}\)....(1)

Let y = e^x … (1)

Differentiating both sides, we get

dy = e^x dx

Substituting in (1),

\(\int\cfrac{dy}{y^2+1}\)

⇒ tan^-1 y

From (1),

⇒ tan^-1 (e^x)

Therefore,

\(\int\cfrac{dx}{(e^x+e^{-x})}=tan^{-1}(e^x)+c\)

Discussion