Chapter A6, Problem 47E

To determine

Toevaluate: the given integrals.

Answer to Problem 47E

The required result is

  $[{\displaystyle \int {\mathrm{sech}}^2}\left(x-\frac{1}{2}\right)\cdot dx=\tanh \left(x-\frac{1}{2}\right)+C$

Explanation of Solution

Given:

  $[{\displaystyle \int {\mathrm{sech}}^2}\left(x-\frac{1}{2}\right)\cdot dx$

Calculation:

Assume $u=\left(x-\frac{1}{2}\right)$

Therefore $du=dx$

Now

  $\begin{array}{l}{\displaystyle \int {\mathrm{sech}}^2}\left(x-\frac{1}{2}\right)dx={\displaystyle \int {\mathrm{sech}}^2}u\cdot du\\ =\tanh u+C\\ =\tanh \left(x-\frac{1}{2}\right)+C\\ \Rightarrow [{\displaystyle \int {\mathrm{sech}}^2}\left(x-\frac{1}{2}\right)\cdot dx=\tanh \left(x-\frac{1}{2}\right)+C\end{array}$

Conclusion:

Therefore, the required result is

  $[{\displaystyle \int {\mathrm{sech}}^2}\left(x-\frac{1}{2}\right)\cdot dx=\tanh \left(x-\frac{1}{2}\right)+C$