Further Maths - Integrating and Differentiating Inverse Trig Functions

Pearson Edexcel Further Mathematics 2022

Created: April 29, 2021 | Read markdown | About these notes | View in context | Study these flashcards |

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Derivatives of inverse trig functions

\[\frac{d}{dx}(\sin^{-1}(x))\]

What is this equal to?

derivative-arcsin
\[\frac{1}{\sqrt{1 - x^2}}\]
\[\frac{d}{dx}(\cos^{-1}(x))\]

What is this equal to?

derivative-arccos
\[-\frac{1}{\sqrt{1 - x^2}}\]
\[\frac{d}{dx}(\tan^{-1}(x))\]

What is this equal to?

derivative-arctan
\[\frac{1}{1 + x^2}\]

Integrals giving inverse trig functions

\[\int \frac{1}{\sqrt{1-x^2}} dx\]

What is this equal to?

integral-to-arcsin
\[\sin^{-1}(x) + c\]
\[\int -\frac{1}{\sqrt{1-x^2}} dx\]

What is this equal to?

integral-to-arccos
\[\cos^{-1}(x) + c\]
\[\int \frac{1}{1 + x^2}dx\]

What is this equal to?

integral-to-arctan
\[\tan^{-1}(x) + c\]

Integrals with a general constant

\[\int \frac{1}{\sqrt{a^2-x^2}} dx\]

What is this equal to?

integral-to-arcsin-with-a
\[\sin^{-1}\left(\frac{x}{a}\right) + c\]
\[\int \frac{1}{a^2 + x^2}dx\]

What is this equal to?

integral-to-arctan-with-a
\[\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + c\]


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