Calculus involving Inverse Trig (Edexcel A Level Further Maths: Core Pure): Revision Note

Differentiating inverse trig functions

What are the inverse trigonometric functions?

• arcsin, arccos and arctan are functions defined as the inverse functions of sine, cosine and tangent respectively

  •  $\arcsin \left(\frac{\sqrt{3}}{2}\right)=\frac{\mathrm{\pi }}{3}$ which is equivalent to $\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$

  •  $\arctan (-1)=\frac{3\pi }{4}$ which is equivalent to $\tan \left(\frac{3\pi }{4}\right)=-1$

What are the derivatives of the inverse trigonometric functions?

• $f\left(x\right)=\arcsin x$

  • $f\text{'}\left(x\right)=\frac{1}{\sqrt{1-x^2}}$

• $f\left(x\right)=\arccos x$

  • $f\text{'}\left(x\right)=-\frac{1}{\sqrt{1-x^2}}$

• $f\left(x\right)=\arctan x$

  • $f\text{'}\left(x\right)=\frac{1}{1+x^2}$

• Unlike other derivatives these look completely unrelated at first

  • their derivation involves use of the identity ${\cos }^{2}x+{\sin }^{2}x\equiv 1$

  • hence the squares and square roots!

• All three are given in the formula booklet

• Note with the derivative of $\arctan x$ that $\left(1+x^2\right)$ is the same as $\left(x^2+1\right)$

How do I show or prove the derivatives of the inverse trigonometric functions?

• For $y=\arcsin x$

  • Rewrite, $\sin y=x$

  • Differentiate implicitly, $\cos y\frac{\mathrm{d}y}{\mathrm{d}x}=1$

  • Rearrange, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\cos y}$

  • Using the identity ${\cos }^{2}y\equiv 1-{\sin }^{2}y$ rewrite, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-{\sin }^{2}y}}$

  • Since, $\sin y=x$, $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}}$

• Similarly, for $y=\arccos x$

  • $\cos y=x$

  • $-\sin y\frac{\mathrm{d}y}{\mathrm{d}x}=1$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sin y}$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-{\cos }^{2}y}}$

  • $\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{\sqrt{1-x^2}}$

• Notice how the derivative of $y=\arcsin x$ is positive but is negative for $y=\arccos x$

  • This subtle but crucial difference can be seen in their graphs

    • $y=\arcsin x$ has a positive gradient for all values of $x$ in its domain

    • $y=\arccos x$ has a negative gradient for all values of $x$ in its domain

Integrating inverse trig functions

How do I integrate inverse trig functions?

• Use integration by parts in the same way you would integrate $\ln x$

• These can be integrated using parts however

  • rewrite as the product ‘$1\times f\left(x\right)$’ and choose $u=f\left(x\right)$ and $\frac{dv}{dx}=1$

  • 1 is easy to integrate and the inverse trig functions have standard derivatives listed in the formula booklet

• The expression $\frac{x}{1+x^2}$ integrates to $\frac{1}{2}\ln \left(1+x^2\right)$

• The expression $\pm \frac{x}{\sqrt{1-x^2}}$ integrates to $\mp \sqrt{1-x^2}$

$\begin{array}{rcl}\int \arcsin \left(x\right)dx& =& x\arcsin \left(x\right)+\sqrt{1-x^2}+c\\ \int \mathrm{arc}\cos \left(x\right)dx& =& x\mathrm{arc}\cos \left(x\right)-\sqrt{1-x^2}+c\\ \int \mathrm{arc}\tan \left(x\right)dx& =& x\mathrm{arc}\tan \left(x\right)-\frac{1}{2}\ln \left(1+x^2\right)+c\end{array}$