Question

How do you evaluate the integral $\int{\dfrac{\sinh x}{1+\cosh x}}$?

Answer 1

Hint: Since, hyperbolic functions are used in the question so, we need to convert them into exponential functions with the help of the formulas $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2},\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$. After that we will use substitution as, $u=2+{{e}^{x}}+{{e}^{-x}}$ and differentiate it with respect to x to solve the question further. Moreover, to get the desired answer we will use $\int{\dfrac{du}{u}}=\log \left| u \right|,\dfrac{d\left( 2 \right)}{dx}=0,\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}},\dfrac{d\left( {{e}^{-x}} \right)}{dx}=-{{e}^{-x}}$.

Complete step by step solution:
Consider the integral function $\int{\dfrac{\sinh x}{1+\cosh x}}dx$.
We will convert this hyperbolic function into exponential function by using the formulas $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2},\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$. Therefore, we get
 \[\begin{align}
  & \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\left( \dfrac{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}}{1+\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{2} \right)} \right)}dx \\
 & \Rightarrow \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\left( \dfrac{\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}}{\dfrac{2+{{e}^{x}}+{{e}^{-x}}}{2}} \right)}dx \\
 & \Rightarrow \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2+{{e}^{x}}+{{e}^{-x}}} \right)}dx\, \\
 & \Rightarrow \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\dfrac{\left( {{e}^{x}}-{{e}^{-x}} \right)dx}{2+{{e}^{x}}+{{e}^{-x}}}}\,...(i) \\
\end{align}\]
Now, we will substitute $u=2+{{e}^{x}}+{{e}^{-x}}$ and differentiate it with respect to x. This results into the following,
\[\begin{align}
  & u=2+{{e}^{x}}+{{e}^{-x}} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{d\left( 2+{{e}^{x}}+{{e}^{-x}} \right)}{dx} \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{d\left( 2 \right)}{dx}+\dfrac{d\left( {{e}^{x}} \right)}{dx}+\dfrac{d\left( {{e}^{-x}} \right)}{dx}\,\,...(i) \\
\end{align}\]
As we know that the differentiation of any constant term is 0 so we have that \[\dfrac{d\left( 2 \right)}{dx}=0\]. Also, the differentiation of ${{e}^{x}}$ is the same. That is, \[\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}\]. Finally, the differentiation of \[\begin{align}
  & \dfrac{d\left( {{e}^{-x}} \right)}{dx}={{e}^{-x}}\times \dfrac{d\left( -x \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( {{e}^{-x}} \right)}{dx}={{e}^{-x}}\times -1 \\
 & \Rightarrow \dfrac{d\left( {{e}^{-x}} \right)}{dx}=-{{e}^{-x}} \\
\end{align}\]
Now, we will substitute these values in equation (ii). Therefore, we get
\[\begin{align}
  & \dfrac{du}{dx}=0+{{e}^{x}}-{{e}^{-x}} \\
 & \Rightarrow du=\left( {{e}^{x}}-{{e}^{-x}} \right)dx \\
\end{align}\]
Keeping the above values in equation (i) we get,
\[\begin{align}
  & \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\dfrac{\left( {{e}^{x}}-{{e}^{-x}} \right)dx}{2+{{e}^{x}}+{{e}^{-x}}}} \\
 & \Rightarrow \int{\dfrac{\sinh x}{1+\cosh x}}dx=\int{\dfrac{du}{u}} \\
\end{align}\]
Since, $\int{\dfrac{du}{u}}=\log \left| u \right|$ so, we get
 \[\begin{align}
  & \int{\dfrac{\sinh x}{1+\cosh x}}dx=\log \left| u \right| \\
 & \Rightarrow \int{\dfrac{\sinh x}{1+\cosh x}}dx=\log \left| 2+{{e}^{x}}+{{e}^{-x}} \right|\left[ \because u=2+{{e}^{x}}+{{e}^{-x}} \right] \\
\end{align}\]
Hence, the value of \[\int{\dfrac{\sinh x}{1+\cosh x}}dx=\log \left| 2+{{e}^{x}}+{{e}^{-x}} \right|\].

Note:
One can think of using the formula \[\cosh x-\sinh x=1\], but it will lead to nowhere. That is why it is better to use conversion of hyperbolic functions into exponential terms. This not only makes the answer easy but also uses differentiation formulas at the basic level. As we can see that, in the solution, we have used substitution for solving the integral. This is done so that we can use the formula $\int{\dfrac{du}{u}}=\log \left| u \right|$ and get the required answer. One should notice all these points carefully and do use a lot of focus to solve this integral. This is because of the fact that integrals may lead to a different answer which is not even required.