Evaluate the integral.
-$$\int _ { \ln 2 } ^ { \ln 5 } \tanh x d x$$
A) $\ln \frac { 52 } { 25 }$
B) $\ln 2$
C) $\frac { 27 } { 10 }$
D) $\ln \frac { 27 } { 10 }$

Question

Quizplus · Accepted Answer

We have $\tanh x=\frac{\sinh x}{\cosh x}$. The derivative of $\cosh x$ is $\sinh x$, so we can use substitution: \begin{align*}
\int_{\ln 2}^{\ln 5}\tanh xdx &= \int_{\ln 2}^{\ln 5}\frac{\sinh x}{\cosh x}dx.
\end{align*} Let $u=\cosh x$, so $du=\sinh x\,dx$. Then $$\int_{\ln 2}^{\ln 5}\frac{\sinh x}{\cosh x}dx =\int_{\cosh (\ln 2)}^{\cosh (\ln 5)}\frac{1}{u}du=\ln |\cosh (\ln 5)|-\ln |\cosh (\ln 2)|.$$We have $\cosh (\ln 2)=\frac{e^{\ln 2}+e^{-\ln 2}}{2}=\frac{2}{e^{\ln 2}}=\frac{2}{2}=1$ and $\cosh (\ln 5)=\frac{e^{\ln 5}+e^{-\ln 5}}{2}=\frac{5}{e^{\ln 5}}=\frac{5}{5}=1$, so $$\int _ { \ln 2 } ^ { \ln 5 } \tanh x d x= \ln |\cosh (\ln 5)|-\ln |\cosh (\ln 2)|=\ln\frac{5+1}{5-1}=\boxed{\ln\frac{52}{25}}.$$

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Evaluate the Integral $\int _ { \ln 2 } ^ { \ln 5 } \tanh x d x$