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

2023
Views:

135

Pages:

4
Author:

Shahbaz A.

calculus shahbaz ahmed August 2024 Proof of relevant formula Z sec θ dθ Z Z sec θ(sec θ + tan θ) dθ secθ + tan θ Z (sec2 θ + sec θ tan θ) dθ secθ + tan θ Z (sec θ tan θ + sec2 θ) dθ secθ + tan θ sec θ dθ = Z sec θ dθ = Z sec θ dθ = Z Z d(sec θ) + d(tan θ) dθ secθ + tan θ sec θ dθ = Z Z sec θ dθ = d(sec θ + tan θ) dθ secθ + tan θ Z sec θ dθ = ln(sec θ + tan θ) + C1 ................................................................ Show that 1 R √ dx x2 −a2 = ln(x + √ x 2 − a2 ) + C Solution Z √ dx x 2 − a2 (1) Putting x = a sec θ in √ x 2 − a2 =⇒ √ √ x2 − a2 = a2 sec2 θ − a2 p √ x2 − a2 = a2 [sec2 θ − 1] p √ x2 − a2 = a2 [tan2 θ] p x2 − a2 = a tan θ Now x = a sec θ dx dθ = d dθ [a sec θ] dx dθ d = a dθ [sec θ] dx dθ = a[sec θ tan θ] 2 (2) dx = a sec θ tan θ dθ (3) Putting equation (2) and equation (3) in equation (1) Z Z dx a sec θ tan θ dθ √ = a tan θ x 2 − a2 Z Z dx √ = sec θ dθ x 2 − a2 Using the formula Z sec θ dθ = ln(sec θ + tan θ) + C1 Z √ dx = ln(sec θ + tan θ) + C1 x 2 − a2 Now x = a sec θ =⇒ x a = sec θ Similarly √ x2 − a2 = a tan θ =⇒ √ x2 −a2 a = tan θ Putting in equation (4) Z dx x √ = ln [ + a x 2 − a2 3 √ x2 − a2 ] + C1 a (4) Z Z √ x+ dx √ = ln[ x 2 − a2 √ x2 − a2 ] + C1 a p dx = ln[x + x2 − a2 ] − ln a + C1 x2 − a2 Putting − ln a + C1 = C √ R dx √ = ln[x + x2 − a2 ] + C x2 −a2 4