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

For $|x|<1$ and all $x$ respectively: $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.$ Proof for arcsin: if $y=\arcsin x$ then $\sin y=x$, so $\cos y\,\frac{dy}{dx}=1$, giving $\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-x^2}}$ (cosine positive on the range of arcsin). The chain rule applies: $\frac{d}{dx}\arcsin(3x)=\frac{3}{\sqrt{1-9x^2}}$ and $\frac{d}{dx}\left[\frac12\arctan\left(x^2\right)\right]=\frac12\cdot\frac{2x}{1+x^4}=\frac{x}{1+x^4}$.

Expressions such as $y=\arcsin x+x\sqrt{1-x^2}$ need the product rule for the second term: $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=\frac{1-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2}=2\sqrt{1-x^2}.$ Combine over a common denominator and use $\frac{1-x^2}{\sqrt{1-x^2}}=\sqrt{1-x^2}$. The result shows that $\int\sqrt{1-x^2}\,dx=\frac12\left(\arcsin x+x\sqrt{1-x^2}\right)+c$.

$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\frac xa+c,\qquad\int\frac{1}{a^2+x^2}\,dx=\frac1a\arctan\frac xa+c.$ The booklet prints the arcsin result and the arctan result in this form. For other coefficients factor out so that the form matches: $\int\frac{1}{\sqrt{9-4x^2}}\,dx=\frac12\int\frac{1}{\sqrt{\frac94-x^2}}\,dx=\frac12\arcsin\frac{2x}{3}+c$. Example: $\int_0^2\frac{1}{4+x^2}\,dx=\frac12\left[\arctan\frac x2\right]_0^2=\frac\pi8$.

When a function is not directly a standard form, choose a substitution that uses a trigonometric identity. For $\sqrt{a^2-x^2}$ use $x=a\sin\theta$: then $dx=a\cos\theta\,d\theta$ and $a^2-x^2=a^2\cos^2\theta$. For $a^2+x^2$ use $x=a\tan\theta$: then $dx=a\sec^2\theta\,d\theta$ and $a^2+x^2=a^2\sec^2\theta$. Change the limits to values of $\theta$ at the same time, and write the integrand entirely in $\theta$ before integrating.

Find $\int_0^1\frac{1}{\left(1+x^2\right)^2}\,dx$ using $x=\tan\theta$. $dx=\sec^2\theta\,d\theta$ and $\left(1+x^2\right)^2=\sec^4\theta$, so the integrand is $\frac{\sec^2\theta}{\sec^4\theta}=\cos^2\theta$. Limits: $x=0\Rightarrow\theta=0$, $x=1\Rightarrow\theta=\frac\pi4$. $\int_0^{\pi/4}\cos^2\theta\,d\theta=\int_0^{\pi/4}\frac12(1+\cos2\theta)\,d\theta=\left[\frac\theta2+\frac{\sin2\theta}{4}\right]_0^{\pi/4}=\frac\pi8+\frac14$.