Inverse Trig Functions

Derivatives:

d/dx[arcsin x] = 1/√(1-x²)

d/dx[arccos x] = -1/√(1-x²)

d/dx[arctan x] = 1/(1+x²)

Corresponding integrals:

∫1/√(1-x²)dx = arcsin x + C

∫1/(1+x²)dx = arctan x + C

With chain rule:

d/dx[arctan(3x)] = 1/(1+(3x)²) · 3 = 3/(1+9x²)

Recognizing the pattern:

When you see 1/(1+x²), think arctan.

When you see 1/√(1-x²), think arcsin.

These integrals can't be done with u-sub or basic rules — you must recognize the pattern.

AP exam: Typically one multiple choice question involves inverse trig.

Reference:

Wikipedia: Inverse Trig

https://en.wikipedia.org/wiki/Inverse_trigonometric_functions