5EAntidiffInvTrig

Antidifferentiation involving Inverse Trig Functions

We know that: math . \qquad \dfrac{d}{dx} \left( \sin^{-1} \left( \dfrac{x}{a} \right) \right) = \dfrac{1}{\sqrt{a^2-x^2}}, \quad x\in \left( -a,a \right) math . math .\qquad \dfrac{d}{dx} \left( \cos^{-1} \left( \dfrac{x}{a} \right) \right) = \dfrac{-1}{\sqrt{a^2-x^2}}, \quad x\in \left( -a,a \right) math . math . \qquad \dfrac{d}{dx} \left( \tan^{-1} \left( \dfrac{x}{a} \right) \right) = \dfrac{a}{a^2+x^2},\quad x\in R math

Hence it follows that we can reverse these to obtain rules for antidifferentiation: math . \qquad \displaystyle {\int} \dfrac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1} \left( \dfrac{x}{a} \right) +c, \quad x \in (-a,a) math

math . \qquad \displaystyle {\int} \dfrac{-1}{\sqrt{a^2-x^2}} dx = \cos^{-1} \left( \dfrac{x}{a} \right) +c, \quad x \in (-a,a) math . math . \qquad \displaystyle {\int} \dfrac{a}{a^2+x^2} dx = \tan^{-1} \left( \dfrac{x}{a} \right) +c, \quad x \in R math . {Note the incorrect formulas for Artan on pages 225, 228 of the __old__ text}

To find the antiderivative involving inverse trig functions, state in the form of one of these three expressions.

Example 1 Find the antiderivative of math . \qquad \displaystyle {\int} \dfrac{-3}{\sqrt{49-x^2}} dx math

Take out 3 in front of the integral sign math \\ . \qquad \displaystyle {\int} \dfrac{-3}{\sqrt{49-x^2}} dx = 3 \displaystyle {\int} \dfrac{-1}{\sqrt{7^2-x^2}} dx \\. \\ . \qquad \qquad \qquad \qquad = 3 \cos^{-1} \left( \dfrac{x}{7} \right) +c, \quad x \in (-7,7) math

Notice that by taking out -3, we could have obtained math . \qquad \qquad \qquad \qquad = -3 \sin^{-1} \left( \dfrac{x}{7} \right) +c, \quad x \in (-7,7) math

Either answer is correct but as a rule, if the function is negative use Arcos.

Example 2 Find the antiderivative of math . \qquad \displaystyle {\int} \dfrac{1}{\sqrt{1-25x^2}} dx math

{Notice the text uses a different (and more complicated) method: See substitution method}

math . \qquad \displaystyle {\int} \dfrac{1}{\sqrt{1-25x^2}} dx = \displaystyle {\int} \dfrac{1}{\sqrt{25 \left( \frac{1}{25} - x^2 \right) }} dx math . math \\ . \qquad \qquad \qquad \qquad = \displaystyle {\int} \dfrac{1}{5 \sqrt{ \left( \frac{1}{5} \right)^2 - x^2 }} dx \\. \\ . \qquad \qquad \qquad \qquad = \dfrac {1}{5} \displaystyle {\int} \dfrac{1}{\sqrt{ \left( \frac{1}{5} \right)^2 - x^2 }} dx math . math \\ . \qquad \qquad \qquad \qquad = \dfrac {1}{5} \sin^{-1} \left( \dfrac {x}{\frac{1}{5}} \right) + c \\. \\ . \qquad \qquad \qquad \qquad = \dfrac {1}{5} \sin^{-1} \left( 5x \right) + c \qquad x \in \left( -\dfrac{1}{5}, \; \dfrac{1}{5} \right) math

Your calculator can find antiderivatives like this example.


 * Example 3 **

Find the antiderivative of: math . \qquad \displaystyle{\int} \dfrac {10}{4 + x^2} \; dx math


 * Solution:**

math . \qquad \displaystyle{\int} \dfrac {10}{4 + x^2} \; dx = 5 \, \displaystyle{\int} \dfrac{2}{2^2 + x^2} \; dx math . math . \qquad \qquad \qquad \quad = 5 \, \tan^{-1} \left( \dfrac {x}{2} \right) + c \qquad x \in R math

**See also** [|www.mathsonline.com.au] Y12Extension --> Functions --> Inverse Functions --> Lesson 5

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