Question

Prove: $\int {\dfrac{{\cosh xdx}}{{\sinh x \times \log \left( {\sinh x} \right)}} = } \log \left( {\log \left( {\sinh x} \right)} \right) + c$

Answer 1

Hint: We can take the LHS and give a substitution for $u = \log \left( {\sinh x} \right)$. Then we can find the derivative of u and substitute back to get an integral with u. Then we can integrate u and resubstitute for u to get the RHS.

Complete step by step answer:

We need to prove that $\int {\dfrac{{\cosh xdx}}{{\sinh x \times \log \left( {\sinh x} \right)}} = } \log \left( {\log \left( {\sinh x} \right)} \right) + c$
We can take the LHS.
\[ \Rightarrow LHS = \int {\dfrac{{\cosh xdx}}{{\sinh x \times \log \left( {\sinh x} \right)}}} \]
We can substitute for $\log \left( {\sinh x} \right)$ as u.
$ \Rightarrow u = \log \left( {\sinh x} \right)$ … (1)
Now we can find the derivative of u with respect to x.
  $ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\log \left( {\sinh x} \right)} \right)$
To find the derivative, we can use the chain rule of differentiation. According to chain rule,
$\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( x \right) \times \dfrac{d}{{dx}}g\left( x \right)$
We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$.Thus, the derivative will become,
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sinh x}} \times \dfrac{d}{{dx}}\left( {\sinh x} \right)$
We know that $\dfrac{d}{{dx}}\left( {\sinh x} \right) = \cosh x$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{\sinh x}} \times \cosh x$
On rearranging, we get,
$ \Rightarrow du = \dfrac{{\cosh x}}{{\sinh x}}dx$… (2)
On substituting (1) and (2) in LHS, we get,
 \[ \Rightarrow LHS = \int {\dfrac{{du}}{u}} \]
We know that \[\int {\dfrac{{du}}{u}} = \log u + c\]
 \[ \Rightarrow LHS = \log u + c\]
Substituting, equation (1), LHS will become,
 \[ \Rightarrow LHS = \log \left( {\log \left( {\sinh x} \right)} \right) + c\]
But we have \[RHS = \log \left( {\log \left( {\sinh x} \right)} \right) + c\]
LHS=RHS
Hence proved.

Note: An alternate method to solve this problem is given by,
We need to prove that $\int {\dfrac{{\cosh xdx}}{{\sinh x \times \log \left( {\sinh x} \right)}} = } \log \left( {\log \left( {\sinh x} \right)} \right) + c$
We know that integrals and derivatives are the opposite operations.
So, to prove this, we can prove that the derivative of the RHS is equal to the term inside the integral of the LHS
Let \[I = \dfrac{d}{{dx}}\log \left( {\log \left( {\sinh x} \right)} \right) + c\]
So, we can take the derivative of the RHS. We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ and by using chain rule, we get,
 \[ \Rightarrow I = \dfrac{1}{{\log \left( {\sinh x} \right)}}\dfrac{d}{{dx}}\log \left( {\sinh x} \right)\]
We know that $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ and again applying chain rule, we get.
 \[ \Rightarrow I = \dfrac{1}{{\log \left( {\sinh x} \right)}} \times \dfrac{1}{{\sinh x}} \times \dfrac{d}{{dx}}\sinh x\]
We know that $\dfrac{d}{{dx}}\left( {\sinh x} \right) = \cosh x$ . So we get,
 \[ \Rightarrow I = \dfrac{1}{{\log \left( {\sinh x} \right)}} \times \dfrac{1}{{\sinh x}} \times \cosh x\]
On rearranging, we get,
 \[ \Rightarrow I = \dfrac{{\cosh x}}{{\sinh x\log \left( {\sinh x} \right)}}\]
This term is equal to the term inside the integral. As integrals and derivatives are opposite operations and I is the derivative of RHS, we can say that the integral of I is equal to the RHS.
LHS=RHS
Hence proved.