Inverse trigonometric functions: differentiation and integrationAQA A-Level Further Maths: Revision notes

$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.$ Derivation for $\arcsin$: let $y=\arcsin x$, so $\sin y=x$ with $-\frac\pi2\le y\le\frac\pi2$. Differentiating, $\cos y\,\frac{dy}{dx}=1$. On this range $\cos y\ge0$, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$ and $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$. The others follow in the same way.

With the chain rule, for a function $u(x)$: $\frac{d}{dx}\arctan u=\frac{u'}{1+u^2}$, and similarly for the others.

Examples: $\frac{d}{dx}\arctan(3x)=\frac{3}{1+9x^2}$ and $\frac{d}{dx}\arccos\frac x2=-\frac{1}{\sqrt{4-x^2}}$. The derivatives of $\arcsin$ and $\arccos$ are not defined at $x=\pm1$ (vertical tangents).

Reading the derivatives backwards, for a constant $a>0$: $\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\frac xa+c,\qquad\int\frac{1}{a^2+x^2}\,dx=\frac1a\arctan\frac xa+c.$ Examples: $\int_0^1\frac{1}{\sqrt{4-x^2}}\,dx=\left[\arcsin\frac x2\right]_0^1=\frac\pi6$ and $\int_0^{\sqrt3}\frac{1}{1+x^2}\,dx=\arctan\sqrt3=\frac\pi3$.

If the coefficient of $x$ is not 1, factorise first: $\int\frac{1}{\sqrt{9-4x^2}}\,dx=\int\frac{1}{2\sqrt{\frac94-x^2}}\,dx=\frac12\arcsin\frac{2x}{3}+c$.

When the integrand contains $\sqrt{a^2-x^2}$, substitute $x=a\sin\theta$, so $\sqrt{a^2-x^2}=a\cos\theta$ and $dx=a\cos\theta\,d\theta$. When it contains $a^2+x^2$, substitute $x=a\tan\theta$, so $a^2+x^2=a^2\sec^2\theta$ and $dx=a\sec^2\theta\,d\theta$. Change the limits at the same time.

Example: $\int_0^1\sqrt{4-x^2}\,dx$ with $x=2\sin\theta$ (limits $0$ to $\frac\pi6$): $\int_0^{\pi/6}4\cos^2\theta\,d\theta=\int_0^{\pi/6}2(1+\cos2\theta)\,d\theta=2\left[\theta+\tfrac12\sin2\theta\right]_0^{\pi/6}=\frac\pi3+\frac{\sqrt3}{2}.$ The $\tan$ substitution also proves the standard result: $\int\frac{1}{a^2+x^2}\,dx=\int\frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta=\frac\theta a=\frac1a\arctan\frac xa$.

A quadratic that is not already $a^2\pm x^2$ can often be made into one by completing the square.

Example: $x^2-4x+13=(x-2)^2+9$, so $\int_2^5\frac{1}{x^2-4x+13}\,dx=\left[\frac13\arctan\frac{x-2}{3}\right]_2^5=\frac{\pi}{12}$.

Example: $5-4x-x^2=9-(x+2)^2$ gives $\int\frac{1}{\sqrt{5-4x-x^2}}\,dx=\arcsin\frac{x+2}{3}+c$.

If the numerator is a multiple of the derivative of the denominator, the integral gives a logarithm; split a general linear numerator into that part plus a constant.

Example: $\frac{x+1}{x^2-4x+13}=\frac{x-2}{x^2-4x+13}+\frac{3}{x^2-4x+13}$.

• $\int_2^5\frac{x-2}{x^2-4x+13}\,dx=\left[\frac12\ln(x^2-4x+13)\right]_2^5=\frac12\ln2$.
• $\int_2^5\frac{3}{x^2-4x+13}\,dx=3\times\frac{\pi}{12}=\frac{\pi}{4}$.

So $\int_2^5\frac{x+1}{x^2-4x+13}\,dx=\frac12\ln2+\frac\pi4$.