# show $\int \frac{1}{\cosh(x)}dx = \arctan(\sinh(x))$ using the substitution $u=\sinh(x)$

I am trying to show that $$\int \frac{1}{\cosh(x)}dx = \arctan(\sinh(x))$$ Using the substitution $$u=\sinh(x)$$

So if $$u=\sinh(x)$$, then $$\frac{du}{dx}=\cosh(x)$$ thus $$\int \frac{1}{\cosh(x)} \times \frac{1}{\cosh(x)} du = \int sech^2(x) du$$ Where do I go from here, the substitution seems to have lead me to no where. Any help would be appreciated.

• $$cosh^2x-sinh^2x=1$$
– Zach
Apr 3 '19 at 19:36

You are almost there. $$\mathrm{sech}^2(x) = 1/\mathrm{cosh}^2(x) = \frac{1}{1+u^2}$$ $$\int \mathrm{sech}^2(x)\mathrm{d}u = \int \frac{1}{1+u^2}\mathrm{d}u = \arctan{(u)} = \arctan{(\sinh(x))}$$
You can use the identity $$\cosh ^2(x)-\sinh ^2(x)=1$$ thus:
$$\int \frac{1}{\cosh(x)}dx =\int \frac{\cosh(x)}{\cosh^2(x)}dx=\int \frac{\cosh(x)}{1+\mathrm{sech}^2(x)}dx$$
and since $$\frac{du}{dx}=\cosh(x)$$
$$\int \frac{1}{\cosh(x)}dx =\int \frac{1}{1+u^2}du=\arctan(u)=\arctan(\sinh(x))$$