Integrating Hyperbolic Functions
Standard integrals of sinh, cosh and the inverse-hyperbolic forms, plus hyperbolic substitution for surds.
Integrating hyperbolic functions
$$\int\sinh x\,dx=\cosh x+c$$
$$\int\cosh x\,dx=\sinh x+c$$
$$\int\frac1{\sqrt{x^2+1}}\,dx=\operatorname{arsinh}x+c$$
$$\int\frac1{\sqrt{x^2-1}}\,dx=\operatorname{arcosh}x+c\;(x>1)$$
$$\int\frac1{1-x^2}\,dx=\operatorname{artanh}x+c\;(|x|<1)$$
Use $\cosh8x\equiv1+2\sinh^24x$, so $\sinh^24x=\dfrac{\cosh8x-1}2$.
$$\int\sinh^24x\,dx=\int\frac{\cosh8x-1}2\,dx=\frac1{16}\sinh8x-\frac x2+c$$
Where a trig substitution would use $x=a\sin u$ (for $\sqrt{a^2-x^2}$), hyperbolic substitutions handle the other two square-root forms: use $x=a\sinh u$ for integrands involving $\sqrt{x^2+a^2}$, and $x=a\cosh u$ for integrands involving $\sqrt{x^2-a^2}$ — exploiting $\cosh^2u-\sinh^2u=1$ to simplify the square root.
$\dfrac{dx}{du}=3\cosh u$, and $x^2+9=9\sinh^2u+9=9\cosh^2u$, so $\sqrt{x^2+9}=3\cosh u$ (taking $\cosh u\geq0$).
$$\int\sqrt{x^2+9}\,dx=\int3\cosh u\cdot3\cosh u\,du=9\int\cosh^2u\,du=9\int\frac{1+\cosh2u}2\,du$$
$$=\frac92\left(u+\frac{\sinh2u}2\right)+c=\frac92u+\frac94\times2\sinh u\cosh u+c=\frac92u+\frac92\sinh u\cosh u+c$$
Back-substitute $u=\operatorname{arsinh}\dfrac x3$, $\sinh u=\dfrac x3$, $\cosh u=\dfrac{\sqrt{x^2+9}}3$:
$$\int\sqrt{x^2+9}\,dx=\frac92\operatorname{arsinh}\frac x3+\frac x2\sqrt{x^2+9}+c$$
Exam-style questions
Question 1 7 marks
Using the substitution $x=2\cosh u$, find $\displaystyle\int\frac1{\sqrt{x^2-4}}\,dx$, for $x>2$, giving your answer in terms of $\operatorname{arcosh}$.
$\dfrac{dx}{du}=2\sinh u$, and $x^2-4=4\cosh^2u-4=4\sinh^2u$, so $\sqrt{x^2-4}=2\sinh u$. (M1, A1)
$$\int\frac1{\sqrt{x^2-4}}\,dx=\int\frac{2\sinh u}{2\sinh u}\,du=\int1\,du=u+c$$ (M1, A1)
Back-substitute $u=\operatorname{arcosh}\dfrac x2$: $\displaystyle\int\frac1{\sqrt{x^2-4}}\,dx=\operatorname{arcosh}\frac x2+c$. (A1, A1, A1)
Question 2 6 marks
Show that $\displaystyle\int_0^{\ln2}\tanh x\,dx=\ln\frac54$.
$\displaystyle\int\tanh x\,dx=\int\frac{\sinh x}{\cosh x}\,dx=\ln(\cosh x)+c$ (since $\cosh x>0$ always). (M1, A1)
$\displaystyle\int_0^{\ln2}\tanh x\,dx=\Big[\ln(\cosh x)\Big]_0^{\ln2}=\ln(\cosh(\ln2))-\ln(\cosh0)$. (M1)
$\cosh(\ln2)=\dfrac{e^{\ln2}+e^{-\ln2}}2=\dfrac{2+\frac12}2=\dfrac54$, and $\cosh0=1$. (A1, A1)
So the integral is $\ln\dfrac54-\ln1=\ln\dfrac54$, as required. (A1)