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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.

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  • $\begingroup$ $$cosh^2x-sinh^2x=1$$ $\endgroup$
    – Zach
    Apr 3 '19 at 19:36
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You are almost there. $\mathrm{sech}^2(x) = 1/\mathrm{cosh}^2(x) = \frac{1}{1+u^2}$ $$ \int \mathrm{sech}^2(x)\mathrm{d}u = \int \frac{1}{1+u^2}\mathrm{d}u = \arctan{(u)} = \arctan{(\sinh(x))} $$

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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 $\frac{du}{dx}=\cosh(x)$

$$\int \frac{1}{\cosh(x)}dx =\int \frac{1}{1+u^2}du=\arctan(u)=\arctan(\sinh(x))$$

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