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

ED
Answered by Emma D. Further Mathematics tutor

2037 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

find all the roots to the equation: z^3 = 1 + i in polar form


Given z=cosx+isinx, show cosx=1/2(z+1/z)


Find the general solution to the differential equation: d^2y/dx^2 - 8 dy/dx +16y = 2x


How do I integrate (sin x)^6?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences