Hyperbolic and trigonometric substitutionsEdexcel International A Level Further Maths: Revision notes

For a constant $a>0$: $\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa+c,\qquad\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+c,$ $\int\frac{dx}{\sqrt{a^2+x^2}}=\operatorname{arsinh}\frac xa+c,\qquad\int\frac{dx}{\sqrt{x^2-a^2}}=\operatorname{arcosh}\frac xa+c\ \ (x>a).$ The two hyperbolic results can also be written $\ln\left(x+\sqrt{x^2+a^2}\right)$ and $\ln\left(x+\sqrt{x^2-a^2}\right)$, up to a constant. Example: $\int_0^3\frac{dx}{9+x^2}=\frac13\arctan1=\frac\pi{12}$, and $\int_0^3\frac{dx}{\sqrt{x^2+9}}=\operatorname{arsinh}1=\ln\left(1+\sqrt2\right)$. Note that only the arctan form has the factor $\frac1a$.

Pick a substitution that turns the awkward expression into a perfect square, using an identity. $\begin{array}{ll}\text{expression}&\text{substitution}\\ a^2+x^2&x=a\tan\theta\ \ (1+\tan^2\theta=\sec^2\theta)\\ \sqrt{a^2-x^2}&x=a\sin\theta\ \ (1-\sin^2\theta=\cos^2\theta)\\ \sqrt{a^2+x^2}&x=a\sinh u\ \ (1+\sinh^2u=\cosh^2u)\\ \sqrt{x^2-a^2}&x=a\cosh u\ \ (\cosh^2u-1=\sinh^2u)\end{array}$ The standard integrals themselves come from these. For $x=a\tan\theta$, $dx=a\sec^2\theta\,d\theta$ and $\int\frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta=\frac\theta a=\frac1a\arctan\frac xa$.

Follow the same steps each time: (1) write $dx$ in terms of the new variable; (2) simplify the integrand using the identity; (3) change the limits for a definite integral; (4) integrate; (5) return to $x$ if the integral is indefinite. Example: $I=\int_0^2\frac{dx}{\left(x^2+4\right)^2}$ with $x=2\tan\theta$. Then $dx=2\sec^2\theta\,d\theta$, $x^2+4=4\sec^2\theta$, and the limits become $0$ to $\frac\pi4$: $I=\int_0^{\pi/4}\frac{2\sec^2\theta}{16\sec^4\theta}\,d\theta=\frac18\int_0^{\pi/4}\cos^2\theta\,d\theta=\frac1{16}\left[\theta+\tfrac12\sin2\theta\right]_0^{\pi/4}=\frac{\pi+2}{64}.$ Use $\cos^2\theta=\frac12(1+\cos2\theta)$ to integrate squares of $\cos\theta$ or $\sin\theta$.

If the surd contains a general quadratic, complete the square first and then use a standard form. $x^2+6x+13=(x+3)^2+4\ \Rightarrow\ \int\frac{dx}{\sqrt{x^2+6x+13}}=\operatorname{arsinh}\frac{x+3}{2}+c.$ Similarly $\int\frac{dx}{x^2+4x+13}=\int\frac{dx}{(x+2)^2+9}=\frac13\arctan\frac{x+2}{3}+c$. For a quadratic with a negative $x^2$ term, such as $5-4x-x^2=9-(x+2)^2$, the result is an $\arcsin$. In a more complicated case the question may give the substitution. For $\int\frac{x^2}{\sqrt{x^2+4}}\,dx$ with $x=2\sinh u$: $dx=2\cosh u\,du$, $\sqrt{x^2+4}=2\cosh u$ and the integrand is $4\sinh^2u\,du$, giving $\sinh2u-2u=\frac12x\sqrt{x^2+4}-2\operatorname{arsinh}\frac x2+c$.

For $\sqrt{1-x^2}$ use $x=\sin\theta$ so that $\sqrt{1-x^2}=\cos\theta$ and $dx=\cos\theta\,d\theta$. $\int_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}\,dx=\int_0^{\pi/6}\sin^2\theta\,d\theta=\left[\tfrac12\theta-\tfrac14\sin2\theta\right]_0^{\pi/6}=\frac\pi{12}-\frac{\sqrt3}{8}.$ The limits convert as $x=\frac12\Rightarrow\sin\theta=\frac12\Rightarrow\theta=\frac\pi6$. The root is positive because $\theta$ is in $\left[0,\frac\pi2\right]$ where $\cos\theta\geq0$. The same approach handles $\int\sqrt{1-x^2}\,dx=\int\cos^2\theta\,d\theta$.