Answer to Question #243086 in Calculus for Akki

Question

Integral sech √x tanh √x/√x dx

Answer 1

Let $I=\int (\frac{sech(\sqrt{x}).tanh(\sqrt{x})}{\sqrt{x}}) dx\\$

Put $\sqrt{x}=t$

$\therefore \frac{1}{2\sqrt{x}}dx=dt\\ \Rightarrow \frac{1}{\sqrt{x}}dx=2.dt$

So, $I=2\int (sech(t).tanh(t))dt\\$

$=2\int (\frac{2}{ e^t + e^{-t}})(\frac{ e^t - e^{-t}}{ e^t + e^{-t}})dt\\ =4\int(\frac{ e^t - e^{-t}}{ (e^t + e^{-t})^2})dt\\$

Put $(e^t + e^{-t})=u$

$\therefore (e^t-e^{-t})dt=du\\$

So, $I=4\int (\frac{1}{u^2})du\\$

$=-\frac{4}{u}+c$ where $c$ is the constant of integration.

$=-\frac{4}{(e^t + e^{-t})}+c\\ =-\frac{4}{(e^{\sqrt{x}} + e^{-\sqrt{x}})}+c\\$