University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

197

Pages:

4
Author:

Shahbaz A.

Calculus shahbaz ahmed August 2024 Show that Z p a2 − x2 a2 −1 x x p 2 dx = a − x2 + C sin + 2 a 2 Solution Z p a2 − x2 dx Putting x = a sin θ =⇒ a2 − x2 = a2 − a2 sin2 θ a2 − x2 = a2 [1 − sin2 θ] a2 − x2 = a2 cos2 θ √ √ a2 − x2 = a2 cos2 θ 1 (1) p a2 − x2 = a cos θ (2) dx = a cos θ dθ (3) Similarly ,differentiating x = a sin θ dx dθ = d dθ [a sin θ] dx dθ d = a dθ [sin θ] dx dθ = a cos θ Putting equation (2) and equation (3) in equation (1) Z p a2 − x2 dx = Z a cos θ(a cos θ )dθ Z p 2 Z a 2 cos2 θ a2 − x2 dx = 2 Z p Z a2 a2 − x2 dx = (1 + cos 2θ )dθ 2 Z Z Z p a2 a2 2 2 a − x dx = dθ + cos 2θ dθ 2 2 Z p a2 a2 sin 2θ 2 2 a − x dx = θ + +C 2 2 2 2 Z p a2 2 sin θ cos θ a2 2 2 +C a − x dx = θ + [ ] 2 2 2 Z p a2 a2 2 2 a − x dx = θ + [ ] sin θ cos θ + C 2 2 x = a sin θ Or x a = sin θ Or sin−1 ( xa ) = θ Similarly x a = sin θ ( xa )2 = sin2 θ ( xa )2 = 1 − cos2 θ 2 2 cos2 θ = 1 − xa2 = a a−x 2 q √ 2 2 cos2 θ = a a−x 2 √ cos θ = 2 a2 −x2 a Putting values of θ, sin θ, cos θ in equation (4) Z p √ 2 2 a x a x a2 − x2 a2 − x2 dx = sin−1 + [ ]( ) +C 2 a 2 a a 3 (4) Z p a2 − x2 a2 −1 x xp dx = sin + [ ] a2 − x2 + C 2 a 2 4