Question

Evaluate the following integral $\int{\dfrac{{{e}^{x}}dx}{\cosh x+ \sinh x}}$. \[\]

Answer 1

Hint: We recall the definitions of integral, integrand, exponential function and hyperbolic function. We use the definition of sine hyperbolic function $\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ and cosine hyperbolic function $\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$ in the integrand and simplify. \[\]

Complete step-by-step solution:
We know that an antiderivative, primitive function or indefinite integral of a function $f$ is a differentiable function $F$ whose derivative is equal to the original function $f$which means${{F}^{'}}=f$. The process of finding integral is called integration and the original function $f$ is called integrand. We write integration with respect to variable $x$as
\[\int{f\left( x \right)}dx=F\left( x \right)+c\]
We know that hyperbolic functions are functions analogous to ordinary trigonometric functions defined for the hyperbola, rather than the circle which is means just $\left( \cos t,\sin t \right)$ with parameter $t$ represents a circle with unit radius the point , the point $\left( \cosh t,\sinh t \right)$ represent form the right half of the equilateral parabola.\[\]
The basic hyperbolic functions are sine hyperbolic function $\left( \sinh x:R\to R \right)$ and cosine hyperbolic function $\left( \cosh x:R\to R \right)$. All the other hyperbolic functions are derived from hyperbolic sine and hyperbolic cosine. \[\]
The hyperbolic sine is defined in terms of exponential function ${{e}^{x}}$ as,
\[\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}....\left( 1 \right)\]
The hyperbolic cosine is defined in terms of exponential function ${{e}^{x}}$ as,
\[\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}....\left( 2 \right)\]
We are asked in the question to evaluate the following integral $\int{\dfrac{{{e}^{x}}dx}{\cosh x+ \sinh x}}$. We see that the integrand the exponential function, the hyperbolic sine, and hyper cosine function is present. We use the definitions of hyperbolic sine (1) and hyperbolic cosine (2) in terms in exponential and put them in the denominator of the integrand we have;
\[\begin{align}
  & \Rightarrow \int{\dfrac{{{e}^{x}}dx}{\cosh x+ \sinh x}} \\
 & \Rightarrow \int{\dfrac{{{e}^{x}}dx}{\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}+\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}}} \\
 & \Rightarrow \int{\dfrac{{{e}^{x}}dx}{\dfrac{{{e}^{x}}+{{e}^{-x}}+{{e}^{x}}-{{e}^{-x}}}{2}}} \\
 & \Rightarrow \int{\dfrac{2{{e}^{x}}dx}{2{{e}^{x}}}} \\
\end{align}\]
We cancel divide $2{{e}^{x}}$ from in the denominator and numerator to have;
\[\Rightarrow \int{dx}=x+c\]
So the integral is $x+c$ where $c$ is real constant of integration.

Note: We can derive other hyperbolic functions using hyperbolic sine and cosine as $\tanh x=\dfrac{{{e}^{2x}}-1}{{{e}^{2x}}+1},\coth x=\dfrac{{{e}^{2x}}+1}{{{e}^{2x}}-1}\left( x\ne 0 \right),\sec hx=\dfrac{2}{{{e}^{x}}+{{e}^{-x}}},\operatorname{cosech}x=\dfrac{2}{{{e}^{x}}-{{e}^{-x}}}$. The integration hyperbolic sine and hyperbolic cosine is given by $\int{\sinh ax}dx=\dfrac{1}{a}\cosh ax+c$ and $\int{\cosh ax}dx=\dfrac{1}{a}\sinh ax+c$ respectively. We can also express $\sinh x,\cosh x$ in terms of series by using exponential series ${{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+...$.