Calculus with hyperbolic functionsAQA A-Level Further Maths: Revision notes

The basic derivatives are $\frac{d}{dx}\sinh x=\cosh x,\qquad\frac{d}{dx}\cosh x=\sinh x,\qquad\frac{d}{dx}\tanh x=\operatorname{sech}^2x.$ The first two follow directly from the exponential definitions. For $\tanh x$ use the quotient rule: $\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=\frac{1}{\cosh^2x}$. The reciprocal functions give $\frac{d}{dx}\operatorname{sech}x=-\operatorname{sech}x\tanh x$, $\frac{d}{dx}\operatorname{cosech}x=-\operatorname{cosech}x\coth x$ and $\frac{d}{dx}\coth x=-\operatorname{cosech}^2x$. Chain, product and quotient rules work as usual, so $\frac{d}{dx}\cosh 2x=2\sinh 2x$.

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: $\int\cosh 2x\,dx=\frac12\sinh 2x+c$ and $\int\operatorname{sech}^2 3x\,dx=\frac13\tanh 3x+c$. Also $\int\tanh x\,dx=\ln\cosh x+c$, because $\tanh x=\frac{\sinh x}{\cosh x}$ and the numerator is the derivative of the denominator. Even powers can be reduced with $\cosh 2x=2\cosh^2x-1$ or $\cosh 2x=1+2\sinh^2x$.

Exact answers use $e^{\ln a}=a$, so $\sinh(\ln a)=\frac{a-\frac1a}{2}$ and $\cosh(\ln a)=\frac{a+\frac1a}{2}$. Example: $\int_0^{\ln2}\cosh 2x\,dx=\frac12\sinh(\ln4)=\frac12\times\frac{15}{8}=\frac{15}{16}$. To find stationary points, solve $\frac{dy}{dx}=0$ and check the range. For $y=2\cosh x-3\sinh x$, $\frac{dy}{dx}=0$ gives $\tanh x=\frac32$, impossible because $|\tanh x|<1$. So the curve has no stationary points. Hyperbolic equations often reduce to $e^x$ quadratics.

Two results (in the formula booklet) are needed: $\int\frac{1}{\sqrt{x^2+a^2}}\,dx=\operatorname{arsinh}\frac xa+c,\qquad\int\frac{1}{\sqrt{x^2-a^2}}\,dx=\operatorname{arcosh}\frac xa+c\ (x>a).$ In logarithmic form, $\operatorname{arsinh}z=\ln\left(z+\sqrt{z^2+1}\right)$ and $\operatorname{arcosh}z=\ln\left(z+\sqrt{z^2-1}\right)$. If the quadratic is not already in the form $x^2\pm a^2$, complete the square. Example: $x^2-6x+5=(x-3)^2-4$, so $\int_5^7\frac{dx}{\sqrt{x^2-6x+5}}=\left[\operatorname{arcosh}\frac{x-3}{2}\right]_5^7=\ln\left(2+\sqrt3\right)$.

Use a hyperbolic substitution when a square root of a quadratic appears:

• $\sqrt{x^2+a^2}$: let $x=a\sinh u$, since $a^2\sinh^2u+a^2=a^2\cosh^2u$.
• $\sqrt{x^2-a^2}$: let $x=a\cosh u$, since $a^2\cosh^2u-a^2=a^2\sinh^2u$. Worked example: $\int_0^4\frac{dx}{\sqrt{x^2+9}}$ with $x=3\sinh u$. Then $dx=3\cosh u\,du$ and $\sqrt{x^2+9}=3\cosh u$, so the integrand is $1$. The limits are $u=0$ and $\sinh u=\frac43$, giving $\cosh u=\frac53$ and $e^u=\sinh u+\cosh u=3$. The answer is $\ln3$. Change the limits with the substitution, or convert back to $x$ at the end. The same substitutions integrate related functions such as $\sqrt{x^2+a^2}$.