Question

Find the value of \[\int {{e^x}\left[ {\log \cosh x - {{\operatorname{sech} }^2}x} \right]dx} \].
A) \[{e^x}\left( {\log \cosh x - \tanh x} \right) + c\]
B) \[{e^x}\log \cosh x + c\]
C) \[ - {e^x}\tanh x + c\]
D) \[{e^x}\left( {\log \cosh x + \tanh x} \right) + c\]

Answer 1

Hint:
Here, we will first apply integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[\int {\left( {{e^x}\log \cosh x} \right)dx} \] of the above equation, where \[f = \ln \cosh x\] and \[g' = {e^x}\] and apply integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[\int {\left( {{e^x}{{\operatorname{sech} }^2}x} \right)dx} \] of the above equation, where \[f = {e^x}\] and \[g' = {\operatorname{sech} ^2}x\]. Then we subtract them to find the required value.

Complete step by step solution:
We are given that
\[\int {{e^x}\left[ {\log \cosh x - {{\operatorname{sech} }^2}x} \right]dx} \]
Rewriting the above equation, we get
\[ \Rightarrow \int {\left( {{e^x}\log \cosh x - {e^x}{{\operatorname{sech} }^2}x} \right)dx} \]
Applying linearity in the above equation, we get
\[ \Rightarrow \int {{e^x}\log \cosh xdx} - \int {{e^x}{{\operatorname{sech} }^2}xdx} \]
Within the above difference, applying integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[\int {\left( {{e^x}\log \cosh x} \right)dx} \] of the above equation, where \[f = \ln \cosh x\] and \[g' = {e^x}\], we get
\[ \Rightarrow \int {{e^x}\ln \cosh xdx} = {e^x}\ln \cosh x - \int {{e^x}\tanh xdx} {\text{ ......eq.(1)}}\]
Within the above difference, applying integration by parts \[\int {fg'} = fg - \int {f'g} \] for \[\int {\left( {{e^x}{{\operatorname{sech} }^2}x} \right)dx} \] of the above equation, where \[f = {e^x}\] and \[g' = {\operatorname{sech} ^2}x\], we get
\[ \Rightarrow \int {{e^x}{{\operatorname{sech} }^2}xdx} = {e^x}\tanh x - \int {{e^x}\tanh xdx} {\text{ ......eq.(2)}}\]
Subtracting the equation (2) from the equation (1), we get
\[
   \Rightarrow \int {{e^x}\log \cosh xdx} - \int {{e^x}{{\operatorname{sech} }^2}xdx} = {e^x}\ln \cosh x - \int {{e^x}\tanh xdx} - \left( {{e^x}\tanh x - \int {{e^x}\tanh xdx} } \right) \\
   \Rightarrow \int {{e^x}\log \cosh xdx} - \int {{e^x}{{\operatorname{sech} }^2}xdx} = {e^x}\ln \cosh x - \int {{e^x}\tanh xdx} - {e^x}\tanh x + \int {{e^x}\tanh xdx} \\
   \Rightarrow \int {\left( {{e^x}\log \cosh x - {e^x}{{\operatorname{sech} }^2}x} \right)dx} = {e^x}\ln \cosh x - {e^x}\tanh x + c \\
 \]

Hence, option A is correct.

Note:
We need to know that while finding the value of indefinite integral, we have to add the constant in the final answer or else the answer will be incomplete. We have to be really thorough with the integrations and differentiation of the functions. The key point in this question is to use the integration by parts \[\int {fg'} = fg - \int {f'g} \] to solve this problem. Do not forget that many integrals can be evaluated in multiple ways and so more than one technique may be used on it, but this problem can only be solved by parts.