Integrating hyperbolic and inverse functionsEdexcel International A Level Further Maths: Revision notes

Reverse the derivatives: $\int\sinh x\,dx=\cosh x+c,\quad\int\cosh x\,dx=\sinh x+c,\quad\int\operatorname{sech}^2x\,dx=\tanh x+c.$ For a linear argument divide by the coefficient: $\int\cosh kx\,dx=\frac1k\sinh kx+c$. Since $\tanh x=\frac{\sinh x}{\cosh x}$ is of the form $\frac{f'}{f}$, $\int\tanh x\,dx=\ln\cosh x+c$ (no modulus needed because $\cosh x>0$). Example: $\int_0^{\ln2}\cosh2x\,dx=\left[\frac12\sinh2x\right]_0^{\ln2}=\frac12\sinh(\ln4)=\frac12\times\frac{4-\frac14}{2}=\frac{15}{16}$.

To integrate $\sinh^2x$ or $\cosh^2x$ use the double-angle identities $\cosh2x=1+2\sinh^2x=2\cosh^2x-1.$ So $\sinh^2x=\frac12\left(\cosh2x-1\right)$ and $\cosh^2x=\frac12\left(\cosh2x+1\right)$. Then $\int\sinh^2x\,dx=\frac14\sinh2x-\frac12x+c$. The identity $\cosh^2x-\sinh^2x=1$ gives $\tanh^2x=1-\operatorname{sech}^2x$, so $\int\tanh^2x\,dx=x-\tanh x+c$.

For a product such as $x\cosh x$, use integration by parts $\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx$ with $u=x$: $\int x\cosh x\,dx=x\sinh x-\int\sinh x\,dx=x\sinh x-\cosh x+c,$ $\int x\sinh x\,dx=x\cosh x-\sinh x+c.$ Example: $\int_0^1x\cosh x\,dx=\left[x\sinh x-\cosh x\right]_0^1=(\sinh1-\cosh1)+1=1-e^{-1}$. Differentiate the polynomial factor, integrate the hyperbolic factor. Use the exponential form $\sinh1-\cosh1=-e^{-1}$ for exact answers.

There is no standard reverse of an inverse function, so write it as $1\times f(x)$ and integrate by parts with $u=f(x)$ and $\frac{dv}{dx}=1$. $\int\operatorname{arsinh}x\,dx=x\operatorname{arsinh}x-\int\frac{x}{\sqrt{1+x^2}}\,dx=x\operatorname{arsinh}x-\sqrt{1+x^2}+c,$ $\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx=x\arctan x-\frac12\ln\left(1+x^2\right)+c.$ The remaining integral is a standard one: $\int\frac{x}{\sqrt{1+x^2}}\,dx$ is $\frac{f'}{\sqrt f}$ form and $\int\frac{x}{1+x^2}\,dx$ is $\frac{f'}{f}$ form. The same method gives $\int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}+c$.

Definite integrals of inverse functions give exact answers in terms of $\pi$ and logarithms. $\int_0^1\arctan x\,dx=\left[x\arctan x-\tfrac12\ln(1+x^2)\right]_0^1=\frac\pi4-\frac12\ln2.$ For $\operatorname{arsinh}$, use $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$. Example: $\int_0^{4/3}\operatorname{arsinh}x\,dx=\frac43\ln3-\frac23$, because $\operatorname{arsinh}\frac43=\ln\left(\frac43+\frac53\right)=\ln3$. A product such as $x\arctan x$ needs $u=\arctan x$ and $\frac{dv}{dx}=x$. Then divide $\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$ and integrate: $\int_0^1x\arctan x\,dx=\left[\tfrac12\left(x^2+1\right)\arctan x-\tfrac12x\right]_0^1=\frac\pi4-\frac12.$