d/dx ar cos x =-1/(squareroot) of 1 - x (squared)

Question

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Answer 1

The question is asking to prove that:

d(arccos(x))/dx≡-1/√(1-x²) which can be written:

arccos(x)≡∫(-1/√(1-x²))dx.

Let x=cosθ, then dx=-sinθdθ=-√(1-x²)dθ, so dθ=-dx/√(1-x²).

Hence ∫(-1/√(1-x²))dx=∫dθ=θ+c where c is constant of integration.

Since θ=arccos(x) because x=cosθ, d(arccos(x))/dx=dθ/dx=-1/√(1-x²) QED