Integrating with Inverse Trigonometric Functions (DP IB Analysis & Approaches (AA): HL): Revision Note

Integrating with inverse trigonometric functions

arcsin, arccos and arctan are (one-to-one) functions defined as the inverse functions of sine, cosine and tangent respectively.

What are the antiderivatives involving the inverse trigonometric functions?

• $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+c$

• $\int \frac{1}{1+x^2}dx=\arctan x+c$

• Note that the antiderivative involving $\arccos x$ would arise from

  
  $\int -\frac{1}{\sqrt{1-x^2}}dx=\arccos x+c$
  

  • However, the negative can be treated as a coefficient of -1 and so
    

    $\int -\frac{1}{\sqrt{1-x^2}}dx=-\int \frac{1}{\sqrt{1-x^2}}dx=-\arcsin x+c$
    

  • Similarly,

$\int \frac{1}{\sqrt{1-x^2}}dx=-\int -\frac{1}{\sqrt{1-x^2}}dx=-\arccos x+c$

• Unless a question requires otherwise, stick to $\arcsin$for integrating (multiples of) $\int \frac{1}{\sqrt{1-x^2}}dx$

How do I integrate these expressions if the denominator is not in the correct form?

• Some problems involve integrands that look very similar to the above

  • but the denominators start with a number other than one

  • There are three particular cases to consider

• The first two cases involve denominators of the form ${\mathit{a}}^{\mathbf{2}}\mathbf{\pm }{\mathbf{\left(}\mathit{b}\mathit{x}\mathbf{\right)}}^{\mathbf{2}}$ (with or without the square root!)

  • In the case $\mathit{b}\mathbf{=}\mathbf{1}$ (i.e. denominator of the form $a^2\pm x^2$) there are two standard results

    • $\int \frac{1}{a^2+x^2}\mathrm{d}x=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+c$

    • $\int \frac{1}{\sqrt{a^2-x^2}}\mathrm{d}x=\arcsin \left(\frac{x}{a}\right)+c,\left|x\right|<a$

    • Note in the first result, $a^2+x^2$ could also be written $x^2+a^2$

      • However $\sqrt{x^2-a^2}$ is not the same as $\sqrt{a^2-x^2}$

  • In cases where $\mathit{b}\mathbf{\ne }\mathbf{1}$ then the integrand can be rewritten by taking a factor of $a^2$

    • the factor will be a constant that can be taken outside the integral

    • the remaining denominator will then start with 1

    • e.g. $9+4x^2=9\left(1+\frac{4}{9}{x}^2\right)=9\left(1+{\left(\frac{2}{3}x\right)}^2\right)$

• The third type of problem occurs when the denominator has a (three term) quadratic

  • i.e.  denominators of the form $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}$
    (a rearrangement of this is more likely)

    • the integrand can be rewritten by completing the square

    • e.g. $5-x^2+4x=5-\left(x^2-4x\right)=5-\left[{\left(x-2\right)}^2-4\right]=9-{\left(x-2\right)}^2$
      This can then be dealt with like the second type of problem above with "$x$" replaced by "$x-2$"

    • This works since the derivative of $x+2$ is the same as the derivative of $x$
      (So there is essentially no reverse chain rule to consider)