A) \[\log (\tan \,h\,x)+c\]
B) \[2\log ({{e}^{x}}+{{e}^{-x}})+c\]
C) \[2\log ({{e}^{x}}-{{e}^{-x}})+c\]
D) \[2\log [\log ({{e}^{x}}+{{e}^{-x}})]+c\]
E) \[\log [\log (\cos \,h\,x)]+c\]
Correct Answer: E
Solution :
Let \[I=\int{\frac{{{e}^{x}}-{{e}^{-x}}}{({{e}^{x}}+{{e}^{-x}})\log (\cosh x)}}dx\] Put \[log(cosh\text{ }x)=t\] \[\Rightarrow \] \[\frac{1}{\cosh x}.\sinh xdx=dt\] \[\Rightarrow \] \[\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}dx=dt\] \[\therefore \]\[I=\int{\frac{1}{t}}dt=\log t+c\]You need to login to perform this action.
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