Chapter A6, Problem 50E

To determine

To Evaluate: the integrals.

Answer to Problem 50E

The result is:

  ${\displaystyle \int \frac{\csc hx\left(\ln t\right).\coth \left(\ln t\right)}{t}.dt}=-\csc hx\left(\ln t\right)+C$

Explanation of Solution

Given:

  ${\displaystyle \int \frac{\csc h\left(\ln t\right)\coth \left(\ln t\right)dt}{t}}$

Calculation:

The given integral is

  ${\displaystyle \int \frac{\csc hx\left(\ln t\right).\coth \left(\ln t\right)}{t}}dt$

assume $u=\ln t$

Therefore $du=\frac{dt}{t}$

Now we have,

  $\begin{array}{l}{\displaystyle \int \frac{\csc hx\left(\ln t\right).\coth \left(\ln t\right)}{t}}dt={\displaystyle \int \csc hxu.\coth u.du}\\ =-\csc hxu+C\\ =-\csc hx\left(\ln t\right)+C\\ \Rightarrow {\displaystyle \int \frac{\csc hx\left(\ln t\right).\coth \left(\ln t\right)}{t}.dt}=-\csc hx\left(\ln t\right)+C\end{array}$

Conclusion:

The result is:

  ${\displaystyle \int \frac{\csc hx\left(\ln t\right).\coth \left(\ln t\right)}{t}.dt}=-\csc hx\left(\ln t\right)+C$