Inverse Trigonometric Functions (AQA A-Level Further Maths): Flashcards

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

$\frac{1}{\sqrt{1-x^2}}$

$-\frac{1}{\sqrt{1-x^2}}$

$\frac{1}{1+x^2}$

Differ only by a negative sign

$\arcsin x + c$

$\arctan x + c$

$x = a\sin u$

$x = a\tan u$

When denominator is quadratic not matching standard form