a) Show that d/dx(arcsin x) = 1/(√ (1-x²)). b) Hence, use a suitable trigonometric substitution to find ∫ (1/(√ (4-2x-x²))) dx.

Will be easier to explain with whiteboard!a) Let y = arcsin x. sin y = x cos y dy/dx = 1 dy/dx = 1/(cos y) dy/dx = 1/(√ (1 - sin²y)) dy/dx = 1/(√( 1 - x²)) as required.b) 4 - 2x - x² = - (x² + 2x - 4) = - [(x+1)² - 5] = 5 - (x+1)² So, ∫ (1/(√ (4-2x-x²))) dx = ∫ (1/(√ (5 - (x + 1)²))) dx Substitution: x + 1 = √ 5 sin θ dx/dθ = √ 5 cos θ dx = √ 5 cos θ dθ So, ∫ (1/(√ (5 - (x + 1)²))) dx = ∫ (1/(√ (5 - 5 sin²θ)) √ 5 cos θ) dθ = ∫ 1 dθ = θ + c = arcsin ((x+1)/√ 5) + c