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

2023
Views:

397

Pages:

3
Author:

Shahbaz A.

Calculus shahbaz ahmed August 2024 Evaluate the integral Z r 1+x dx 1−x Z r 1+x dx 1−x Solution Let x = sin θ Therefore 1+x 1−x = 1+sin θ 1−sin θ 1+x 1−x = 1+sin θ 1−sin θ 1+x 1−x = (1+sin θ)2 1−sin2 θ 1+x 1−x = (1+sin θ)2 cos2 θ × 1+sin θ 1+sin θ 1+x 1−x θ 2 = [ 1+sin cos θ ] q q 1+x θ 2 [ 1+sin 1−x = cos θ ] 1 (1) r 1 + x 1 + sin θ = 1−x cos θ (2) Also x = sin θ Taking derivative with respect to θ dx dθ = d dθ (sin θ) dx dθ = cos θ dx = cos θ dθ (3) Putting equation (2) and equation (3) in equation (1) Z r 1+x dx = 1−x Z 1 + sin θ cos θ dθ cos θ Z r Z 1+x dx = (1 + sin θ) dθ 1−x Z r Z Z 1+x dθ + sin θ dθ dx = 1−x Z r 1+x dx = θ − cos θ + C 1−x 2 (4) Now x = sin θ =⇒ sin−1 x = θ Similarly x = sin θ x2 = sin2 θ x2 = 1 − cos2 θ cos2 θ = 1 − x2 √ √ cos2 θ = 1 − x2 √ cos θ = 1 − x2 Putting the values of θ and cos θ in equation (4) Z r p 1+x dx = sin−1 x − 1 − x2 + C 1−x 3