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

2023
Views:

139

Pages:

4
Author:

Shahbaz A.

Integration 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 4 + 2x + x2 Solution 2 Z x dx 4 + 2x + x2 (1) Now x 4+2x+x2 = 1 2x+2−2 2 4+2x+x2 x 4+2x+x2 = 1 2x+2 2 4+2x+x2 − 1 2 2 4+2x+x2 x 4+2x+x2 = 1 2x+2 2 4+2x+x2 − 1 (x2 +2x+1)+3 xdx 4+2x+x2 = 1 d(x2 +2x+4) 2 4+2x+x2 Now − dx √ (x+1)2 +( 3)2 Z Z xdx = 4 + 2x + x2 Z xdx 1 = 4 + 2x + x2 2 Z 1 d(x2 + 2x + 4) dx √ − 2 4 + 2x + x2 (x + 1)2 + ( 3)2 d(x2 + 2x + 4) − x2 + 2x + 4 Using the formulas R du u = ln u R du 1 −1 u u2 +a2 = a tan a 3 Z dx √ (x + 1)2 + ( 3)2 Z 1 1 xdx 2 −1 x + 1 √ √ +C = ln |x + 2x + 4| − tan 4 + 2x + x2 2 3 3 4