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

2023
Views:

404

Pages:

3
Author:

Shahbaz A.

Integration 18 shahbaz ahmed August 2024 Relevant concepts and formulas limn−→∞ (1 + n1 )n = e Putting x= 1 n When n −→ ∞ ⇐⇒ x → 0 Also x= 1 n ⇐⇒ n = 1 x When x → 0 ⇐⇒ n → ∞ So 1 limx→0 (1 + x) x = e Where e is an irrational number such that 2 < e < 3 Also loge x = ln x 1 d dx (ln x) R 1 x = 1 x dx = ln |x| + C Also R du u2 +a2 Putting u = a tan θ du = a sec2 θ dθ u2 + a2 = a2 tan2 θ + a2 = a2 (1 + tan2 θ) = a2 sec2 θ R du R a sec2 θ dθ = 2 2 u +a a2 sec2 θ R du R 1 = dθ = a1 θ + C u2 +a2 a Since u = a tan θ u a = tanθ tan−1 ua = θ R du 1 −1 u u2 +a2 = a tan a +C ......................................... Evaluate the integral Z x dx x4 + 2x2 + 5 Solution 2 Z xdx = x4 + 2x2 + 5 Z 1 2x dx [ 4 ] 2 x + 2x2 + 5 Since x4 + 2x2 + 5 = x4 + 2x2 + 1 + 4 = (x2 + 1)2 + 22 . Therefore Z xdx 1 = x4 + 2x2 + 5 2 Z [ 2x dx ] (x2 + 1)2 + 22 Also d(x2 + 1) = 2xdx Z xdx 1 = x4 + 2x2 + 5 2 Z d(x2 + 1) [ 2 ] (x + 1)2 + 22 Using the formula R du 1 −1 u u2 +a2 = a tan a +C Z 2 1 xdx −1 x + 1 = tan [ ]+C x4 + 2x2 + 5 4 2 3