Further integration and reduction formulae

What you’ll learn

• Choose trigonometric or hyperbolic substitutions.
• Integrate inverse functions and derive reduction relations.

Understand the idea

Quadratic surds suggest substitutions that match an identity: use $x=a\sin t$ for $\sqrt{a^2-x^2}$ and $x=a\sinh t$ for $\sqrt{a^2+x^2}$. Complete the square first when necessary, change the differential and limits, and keep track of the sign of any square root.

Integrals of inverse functions often yield to integration by parts with the inverse function as $u$ and $dv=dx$. A reduction formula links an integral with index $n$ to one with a lower index; derive it with integration by parts and an identity, retain boundary terms, and finish using a known starting integral.

Key formulae

$$\int\frac{dx}{\sqrt{a^2+x^2}}=\operatorname{arsinh}(x/a)+C\quad(a>0)$$

$$\int\frac{dx}{a^2+x^2}=\frac1a\arctan(x/a)+C\quad(a>0)$$

$$\int u\,dv=uv-\int v\,du$$

$$I_n=\int_0^{\pi/2}\sin^nx\,dx,\qquad I_n=\frac{n-1}{n}I_{n-2}\quad(n\ge2)$$

Worked example

Find $\int\operatorname{arctan}x\,dx$.

1. Choose $u=\arctan x$ and $dv=dx$.
2. Integration by parts gives $x\arctan x-\int x/(1+x^2)\,dx$.
3. In the remaining integral use $w=1+x^2$, so $dw=2x\,dx$.

Answer: $x\arctan x-\frac12\ln(1+x^2)+C$.