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

2023
Views:

134

Pages:

3
Author:

Shahbaz A.

integration shahbaz ahmed August 2024 Evaluate the integral Z √ dx 7 − 6x − x2 √ dx 7 − 6x − x2 Solution Z 7 − 6x − x2 = −x2 − 6x + 7 7 − 6x − x2 = −(x2 + 6x − 7) 7 − 6x − x2 = −(x2 + 6x + 9 − 9 − 7) 7 − 6x − x2 = −[(x2 + 6x + 9) − 9 − 7] 7 − 6x − x2 = −[(x + 3)2 − 16] 7 − 6x − x2 = [16 − (x + 3)2 ] 7 − 6x − x2 = [42 − (x + 3)2 ] Let x + 3 = 4 sin θ 1 (1) Or (x + 3)2 = 42 sin2 θ 7 − 6x − x2 = [42 − 42 sin2 θ] 7 − 6x − x2 = 42 [1 − sin2 θ] 7 − 6x − x2 = 42 cos2 θ √ √ 7 − 6x − x2 = 42 cos2 θ p 7 − 6x − x2 = 4 cos θ (2) Also x + 3 = 4 sin θ d dθ [4 sin θ] d dθ (x + 3) = dx dθ + d dθ (3) dx dθ + 0 = 4 cos θ dx dθ = 4 cos θ d = 4 dθ [sin θ] dx = 4 cos θ dθ Putting equation (2) and equation (3) in equation (1) 2 (3) Z dx √ = 7 − 6x − x2 Z Z Z 4 cos θ dθ 4 cos θ dx √ = 7 − 6x − x2 √ Z dθ dx =θ+C 7 − 6x − x2 Now x + 3 = 4 sin θ x+3 4 = sin θ sin−1 x+3 4 =θ Hence Z √ dx x+3 = sin−1 [ ]+C 4 7 − 6x − x2 3