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

2023
Views:

345

Pages:

2
Author:

Shahbaz A.

Calculus shahbaz ahmed August 2024 Proof of relevant formula R du u = | ln u| + c1 d(cot u) = − csc2 u du d(csc u) = − csc u cot u du d(csc u) − d(cot u) = d(csc u − cot u) R csc u du R csc u du = R csc u(cscu−cot u) csc u−cot u du R csc u du = R csc2 u−csc u cot u csc u−cot u du R csc u du = R csc2 u−csc u cot u csc u−cot u du R csc u du = R − csc u cot u+csc2 u csc u−cot u R csc u du = R d(csc u)−d(cot u) csc u−cot u R csc u du = R d(csc u−cot u) csc u−cot u R csc u du = ln|csc u − cot u| + c2 du 1 ................................................... Evaluate the integral Z dx 1 2 √ 3 2 sin x + cos x Solution Z dx 1 2 Z √ sin x + 3 2 cos x Z dx dx 3 1 sin x cos π3 + cos x sin π3 2 sin x + 2 cos x Z Z dx dx √ = 1 sin (x + π3 ) sin x + 3 cos x = √ 2 Z 2 Z dx = √ 1 2 sin x + 3 2 cos x π csc (x + ) dx 3 Using the formula R csc u du = ln|csc u − cot u| Z dx π π = ln |csc (x + ) − cot (x + )| + C 3 1 3 3 2 sin x + 2 cos x √ 2