Integrals giving inverse trigonometric functions

The standard integrals that give inverse sine and inverse tangent, evaluating definite integrals exactly in terms of π, adjusting coefficients such as 1/(4 + 9x²), and areas under these curves.

The standard forms

Standard integrals

$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C \qquad \int \frac{dx}{a^2 + x^2} = \frac1a\tan^{-1}\frac{x}{a} + C$$

These follow at once from the derivatives in the last section. (The integral of $\dfrac{-1}{\sqrt{a^2 - x^2}}$ is $\cos^{-1}\dfrac{x}{a} + C$, but $-\sin^{-1}\dfrac{x}{a} + C$ is just as correct.)

Example 1●

Evaluate $\displaystyle\int_0^1 \frac{dx}{1 + x^2}$ and $\displaystyle\int_0^{\frac32} \frac{dx}{\sqrt{9 - x^2}}$.

$\Big[\tan^{-1}x\Big]_0^1 = \dfrac{\pi}{4}$.

$\Big[\sin^{-1}\dfrac{x}{3}\Big]_0^{\frac32} = \sin^{-1}\tfrac12 = \dfrac{\pi}{6}$.

Area under y = 1/(1 + x²)

As b grows, the area approaches 2 × π/2 = π ≈ 3.14, even though the region is infinitely wide.

Try this

How large must $b$ be for the area to exceed $3$?

Adjusting the coefficient

Example 2●●●

Find $\displaystyle\int \frac{dx}{4 + 9x^2}$.

Take out $9$: $\dfrac19 \displaystyle\int \frac{dx}{\frac49 + x^2}$, which has $a = \tfrac23$.

$= \dfrac19 \times \dfrac32 \tan^{-1}\dfrac{3x}{2} + C = \dfrac16\tan^{-1}\dfrac{3x}{2} + C$.

Flashcards

$displaystyleint rac{dx}{sqrt{a^2 - x^2}}$

$\sin^{-1}\dfrac{x}{a} + C$

$displaystyleint rac{dx}{a^2 + x^2}$

$\dfrac1a\tan^{-1}\dfrac{x}{a} + C$

$displaystyleint_0^1 rac{dx}{1 + x^2}$

$\dfrac{\pi}{4}$

Practice

Evaluate $\displaystyle\int_0^1 \frac{dx}{1 + x^2}$. Type pi for $\pi$.
Evaluate $\displaystyle\int_0^{\frac32} \frac{dx}{\sqrt{9 - x^2}}$.
Evaluate $\displaystyle\int_0^2 \frac{dx}{4 + x^2}$.
Evaluate $\displaystyle\int_0^{\sqrt3} \frac{dx}{1 + x^2}$.
Evaluate $\displaystyle\int_{-1}^{1} \frac{dx}{1 + x^2}$.
Evaluate $\displaystyle\int_0^{2} \frac{dx}{\sqrt{4 - x^2}}$.
Evaluate $\displaystyle\int_0^{\frac12} \frac{dx}{\sqrt{1 - 4x^2}}$.
$\displaystyle\int \frac{dx}{4 + 9x^2} = $
$\displaystyle\int \frac{dx}{\sqrt{25 - x^2}} = $