Title CBSE - MATHEMATICS (PART - II).pdf ✅

S.No. Content Page 01. Chapter-7 : Integrals 01 - 46 02. Chapter-8 : Application of Integrals 47 - 68 03. Chapter-9 : Differential Equations 69 – 96 04. Chapter-10 : Vector Algebra 97 – 130 05. Chapter-11 : Three Dimensional Geometry 131 – 156 06. Chapter-12 : Linear Programming 157 – 176 07. Chapter-13 : Probability 177 - 216 08. CBSE Question Paper & Solution – 2022-23 217 – 240 09. CBSE Sample Paper & Solution – 2023-24 241 - 266

2 E (viii) d dx (sin x) = cos x cos x dx sin x C    (ix) d dx (tan x) = sec2 x 2 sec x dx tan x C    (x) d dx (–cot x) = cosec2 x 2 cosec x dx cot x C     (xi) d dx (sec x) = sec x tan x sec x tan x dx sec x C    (xii) d dx (–cosec x) = cosec x cot x cosecx cot x dx cosecx C     (xiii) d dx (sin–1 x) = 2 1 1 x  1 2 1 dx sin x C 1 x      (xiv) d dx (–cos –1 x) = 2 1 1 x  1 2 1 dx cos x C 1 x       (xv) d dx (tan–1 x) = 2 1 1 x  1 2 1 dx tan x C 1 x      (xvi) d dx (–cot–1 x) = 2 1 1 x  1 2 1 dx cot x C 1 x       (xvii) d dx (sec–1 x) = 2 1 x x 1 1 2 1 dx sec x C x x 1      (xviii) d dx (–cosec –1 x) = 2 1 x x 1 1 2 1 dx cosec x C x x 1       4. BASIC THEOREMS ON INTEGRATION : If f(x), g(x) are two functions of a variable x and k is a constant, then (i) k f (x)dx k f (x)dx    (ii) [f (x) g(x)]dx f (x)dx g(x)dx       (iii) d dx   f (x)dx f (x)   (iv) d f(x) dx f(x) dx         + C (v) If f x   dx = (x) + C, then f ax b     dx = 1 a (ax + b) + C   d 1 (ax b) '(ax b) f (ax b) dx a             
E 3 Illustration 1: Evaluate the following integrals : (i) 1 1 cos 2x tan dx 1 cos 2x           (ii) 1 1 x x dx x x                (iii) 2 2x dx (2x 1)   (iv) sin 5x sin 3x dx  (v) cos 2x cos 2 dx cos x cos      Solution: (i) Let I = 2 1 1 2 1 cos 2x 2sin x tan dx tan dx 1 cos 2x 2cos x                    =  tan–1 (tan x) dx (We are assuming that 0 < x < 2  ) = 1 2 x dx x C 2    . (ii) Let I = 1 1 x x 1 x x dx x x dx x x x x x x                           = 3/2 1/2 1/2 –3/2 x dx x dx x dx x dx         = 2 2 5/2 3/2 2 x x 2 x C 5 3 x     (iii) Let I =   2 2 2 2x (2x 1) 1 1 1 dx dx dx (2x 1) (2x 1) 2x 1 (2x 1)                   1 log 2x 1 (2x 1) C 2 ( 1) 2         1 1 log 2x 1 C 2 (2x 1)            (iv) Let I = sin 5x sin 3x dx   1 2 2sin 5x sin 3x  dx = 1 2 cos 2x – cos 8x    dx = 1 2 sin 2x sin8x C 2 8         = 1 16 (4sin 2x – sin 8x) + C. (v) Let I = 2 2 cos 2x cos 2 (2cos x 1) (2cos 1) dx cos x cos cos x cos              dx = 2  (cos x + cos ) dx = 2 sin x + 2x cos  + C Illustration 2: If f '(x) = x2 – 2 1 x and f(1) = 1 3 , find f(x) Solution: Given f '(x) = x2 – 2 1 x 2 2 1 f (x) x dx x          [Integration on both sides]  f(x) 2 –2  (x – x )dx  3 1 x x 1 1 3 C x C 3 1 3 x         . Also f(1) = 1 3  1 3 × 13 + 1 1 + C = 1 3  1 3 + 1 + C = 1 3 C = –1 f(x) = 1 3 x 3 + 1 x – 1