how do we prove integral sechx?

Question

$$\int \text{sech} x\, dx = 2\arctan(\tanh x/2) $$

how do we prove this in step by step process??

Is $\arctan(\sinh x)$ equal to $2\arctan(\tanh x/2)$ ??

Answer 1

To obtain the result you are after one can use the equivalent of a Weierstrass substitution for hyperbolic functions (a so-called hyperbolic tangent half-argument substitution). In this case let $t = \tanh \frac{x}{2}$. For such a substitution it can be readily shown that in terms of $t$ we have: $$\sinh x = \frac{2t}{1 - t^2}, \quad \cosh x = \frac{1 + t^2}{1 - t^2}, \quad dx = \frac{2}{1 - t^2} \, dt.$$ So under such a substitution your integral becomes \begin{align*} \int \text{sech} x \, dx &= \int \frac{1}{\cosh x} \, dx\\ &= \int \frac{1 - t^2}{1 + t^2} \cdot \frac{2}{1 - t^2} \, dt\\ &= 2 \int \frac{dt}{1 + t^2}\\ &= 2 \tan^{-1} (t) + C\\ &= 2 \tan^{-1} \left (\tanh \frac{x}{2} \right ) + C. \end{align*}

Answer 2

$$\newcommand{\sech}{\operatorname{sech}} \begin{align} \int\sech(x)\,\mathrm{d}x &=\int\sech^2(x)\,\mathrm{d}\sinh(x)\tag1\\ &=\int\frac1{1+\sinh^2(x)}\,\mathrm{d}\sinh(x)\tag2\\[3pt] &=\arctan(\sinh(x))+C\tag3\\[12pt] &=\arctan\left(2\sinh(x/2)\cosh(x/2)\right)+C\tag4\\[6pt] &=\arctan\left(\frac{2\tanh(x/2)}{\sech^2(x/2)}\right)+C\tag5\\ &=\arctan\left(\frac{2\tanh(x/2)}{1-\tanh^2(x/2)}\right)+C\tag6\\[6pt] &=2\arctan(\tanh(x/2))+C\tag7 \end{align} $$ Explanation:
$(1)$: $\mathrm{d}\sinh(x)=\cosh(x)\,\mathrm{d}x$
$(2)$: $\cosh^2(x)=\sinh^2(x)+1$
$(3)$: standard arctan integral
$(4)$: hyperbolic duplication formula
$(5)$: $\sinh(x)=\tanh(x)\cosh(x)$
$(6)$: $\sech^2(x)=1-\tanh^2(x)$
$(7)$: $\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$