Maths 12th-part-2 PDF

CHAPTER 7 Integrals Page 239 CHAPTER 7...

CHAPTER 7 Integrals Page 239 CHAPTER 7 INTEGRALS OBJECTIVE QUESTIONS = #0 1 tdt : #a f ^x h dx = - #b f ^x h dx D b a t2 1 = :2D dx 1. The integral # is equal to : 0 9 - 4x2 1 = b 2 - 0l = 1 (a) 1 sin-1 c 2x m + c (b) 1 sin-1 c 2x m + c 2 6 3 2 3 Thus (s) is correct option. 2 x (c) sin-1 c m + c (d) sin-1 c x m + c 3 2 3 2 3 3. Anti-derivative of tan x - 1 with respect to x is Sol : OD 2024 tan x + 1 p (a) sec2 a p - x k + c (b) - sec2 a 4 - x k + c 1 4 I = #dx (c) log sec a p - x k + c (d) - log sec a p - x k + c 9 - 4x2 4 4 = # 1 dx Sol : OD 2023 9 ^1 - 43 x2h tan x - 1 dx = # 1 dx I = tan x + 1 # 3 1 - ^ 23 x h2 = - # 1 - tan x dx 1 + tan x =1# 1 dx 3 1 - _ 23 x i = - # tan a p - x k dx 2 4 sin-1 _ 23 x i =1 +C p = ln sec a - x k + c 3 2 3 4 Thus (c) is correct option. = 1 sin-1 c 2 x m + c 2 3 Thus (b) is correct option. 4. If d dx f ^x h = 2x + x3 and f ^1 h = 1, then f ^x h is p/2 (a) x 2 + 3 log | x |+ 1 (b) x 2 + 3 log | x | 2. The value of #p/4 cot q cosec2 qdq is (c) 2 - 32 (d) x 2 + 3 log | x |- 4 (a) 1 (b) - 1 x 2 2 Sol : OD 2023 (c) 0 (d) - p 8 We have dx ^ h d f x = 2x + 3 x Sol : OD 2024 p/2 Integrating on both sides we have I = #p/4 cot q cosec2 qdq...(i) d f (x) dx # dx = # b2x + x3 ldx Let cot q = t then we have f ^x h = x 2 + 3 ln | x |+ c...(1) 2 - cosec qdq = dt Since f ^1 h = 1, substituting x = 1 in above equation cosec2 qdq = - dt we have When q = p t = cot p = 1 f ^1 h = 1 2 + 3 ln | 1 |+ C 4 4 When q = p p t = cot = 0 1 = 1+0+C & C = 0 2 2 Hence, given integral become Thus f ^x h = x2 + 3 ln | x | 0 Thus (b) is correct option. I = - # tdt 1 CHAPTER 7 Integrals Page 241 P =1 = 2# 1 = 2 4 2 xe - 1 + ex - 1 dx is equal to Thus (c) is correct option. 11. # xe + ex 3a # b ax - 1 l dx is equal to 1 2 14. (a) log (xe + ex ) + C (b) e log (xe + ex ) + C 0 a-1 (c) 1 log (xe + ex ) + C (d) None of these (a) a - 1 + (a - 1) -2 (b) a + a-2 e Sol : OD 2011 (c) a - a2 (d) a2 + 12 a xe - 1 + ex - 1 dx Sol : SQP 2020 We have I = xe + ex # 1 ax - 1 2 (ax - 1) 3 b a - 1 l dx = (a 1) 2 ; 3a E 1 1 Substituting xe + ex = t & e (xe - 1 + ex - 1) dx = dt 3a # 0 - 3 # a 0 I = 1 # dt = 1 log t + C 1 e t e = [(a - 1) 3 + 1] (a - 1) 2 = 1 log (xe + ex ) + C e = (a - 1) + (a - 1) -2 Thus (c) is correct option. Thus (a) is correct option. 12. # dx is equal to 1 dx is x (x7 + 1) 15. The value of # 0 ex + e 7 7 (a) log c 7x m + C (b) 1 log c 7x m + C (a) 1 log b 1 + e l (b) log b 1 + e l x +1 7 x +1 e 2 2 7 7 (c) log c +7 1 m + C x (d) log c x +7 1 m + C 1 1 (c) log (1 + e) (d) log b 2 l x 7 x e 1+e Sol : Comp 2017, Delhi 2008 Sol : Foreign 2018, Delhi 2010 dx #0 1 We have I = # x (xdx+ 1) 7 We have ex + e I = 1 dx Substituting x7 = t & dx = 1 6 dt we have = 0 ex 1 + e # 7x a ex k 1 dt e I = 7 t (t + 1) # Substituting 1 + x = t e e 0 - x dx = dt & 1x dx = - 1 dt we obtain = 1 # b 1 - 1 l dt e 7 t t+1 e e 1 2 1 = 1 [log t - log (t + 1)] + C I =- # e 1+e t dt 7 = 1 log b t l + C = - 1 [log t] 12 + e 7 t+1 e 7 - = 1 [log 2 - log (1 + e)] = 1 log c 7x m + C e 7 x +1 - = 1 log b 2 l Thus (b) is correct option. e 1+e = log a 2 k 1 1 +e p/2 13. #0 cos x2 dx is equal to e (a) 1 (b) - 2 Thus (a) is correct option. (c) 2 (d) 0 2 16. The value of #-2 (x cos x + sin x + 1) dx is Sol : Foreign 2009 (a) 2 (b) 0 (c) - 2 (d) 4 cos a x k dx = cos a x k dx p/2 p/2 #0 2 #0 2 Sol : Delhi 2015 = 2 9sin a x kC ] p/2 2 0 2 I = #-2 (x cos x + sin x + 1) dx = 2 8sin p - sin 0 B 4 2 2 2 = #-2 x cos x dx + #-2 sin x dx + #-2 1 dx = 2 sin p 4 CHAPTER 7 Integrals Page 243 a -1/2 + 1 (a) 2 #0 f (x) dx (b) 0 = x - 1/2 + 1 +k (c) 1 (d) - 1 1/2 = x +k Sol : Comp 2016 1/2 = 2 x +k We have f ^x h = - f ^x h Thus (b) is correct option. Thus f ^x h is odd function and for odd function b a 26. # x5 dx = # f (x) dx = = 0 a 6 6 (b) b - a -a Thus (b) is correct option. (a) b5 - a5 6 6 6 b a (c) a - b (d) a5 - b5 23. #a f (x) dx + # b f (x) dx = 6 b Sol : Delhi 2007 (a) 1 (b) 2 # f (x) dx a b (c) - 2 # a f (x) dx (d) 0 Let, I = #a x5 dx b 6 b Sol : OD 2012, Delhi 2010 =x 6 a b f ^x h dx we have a Using property # f^x hdx =- # 6 = b -a 6 b a 6 6 b a 6 6 I = # f (x) dx + # f (x) dx =b -a a b 6 f ^x h dx b b = # f ^x h dx - # Thus (b) is correct option. a a =0 1 (tan-1 x) 2 27. #0 1 + x2 dx = Thus (d) is correct option. 3 (a) 1 (b) p 1 64 24. # (x) dx = (c) p2 (d) None of these 0 192 (a) 0 (b) 1 Sol : Delhi 2010 (c) 2 (d) 1 2 1 (tan-1 x) 2 Sol : Comp 2016 Let, I = # 0 1 + x2 dx 1 Let tan-1 x = t & 1 dx = dt We have I = # xdx 0 1 + x2 2 1 When x = 0 , t =0 =x 2 t =p 0 and x = 1, 4 =1-0 =1 2 2 2 1 (tan-1 x) 2 Thus (d) is correct option. Thus I = # 0 1 + x2 dx p 4 dx = 25. # x = #t 2 dt 0 3 p 4 (a) x +k (b) 2 x + k =t 3 0 (c) x + k (d) 2 x3/2 + k = 1a p k - 0 3 3 3 4 3 Sol : OD 2008 p 3 = 192 We have I = # dxx Thus (d) is correct option. = #x -1/2 dx CHAPTER 7 Integrals Page 245 = # cot q cosec qdq 37. # sin1 x++sincos2xx dx is equal to = - cos ec q + C (a) sin x (b) x Thus (b) is correct option. (c) cos x (d) tan x Sol : SQP 2020 34. # cosec x cot xdx is (a) - cosec x + c (b) - cot x # sin1 x++sincos2xx dx = # sin x + cos x (sin x + cos2 x) + 2 sin x cos x 2 dx (c) cosec x (d) None sin x + cos x dx = # (sin x + cos x) 2 Sol : Comp 2008 sin x + cos x dx I = # cosec x cot xdx = # sin x + cos x = - cosec x +c = # dx =x d x dx ^ h =- cosec x + c Thus (b) is correct option. p/2 Thus (a) is correct option. 38. The value of # log cos x dx 0 is equal to the value of: p/2 p/2 35. # sin x x dx is equal to (a) # log sin x dx 0 (b) # log sec x dx 0 (a) cos x (b) - cos x p/2 p/2 (c) # log cos x dx (d) # log tan x dx (c) 2 cos x (d) - 2 cos x 0 0 Sol : OD 2012 Sol : OD 2007 p/2 We have I = # sin x dx We have I = # log cos x dx 0 x b b Putting x =t & 1 dx = dt we have Using property # f (x) $ dx = # f (a + b - x) $ dx we get a a 2 x p/2 I = # log cos a x - x k dx I = # 2 sin tdt 0 2 p/2 = - 2 cos t + c = # log sin x dx 0 = - 2 cos x + c Thus (a) is correct option. Thus (d) is correct option. 2 39. I= # | 1 - x | dx 0 is equal to 36. # log 2x dx is equal to (a) 0 (b) 1 2 (a) x log x - 1 (b) x log 2x + 1 (c) 1 (d) None of these (c) x log 2x - x (d) x log 2x + 2x Sol : Delhi 2007 Sol : Foreign 2018, OD 2013 We have | 1 - x | = 1 - x , if 0 # x # 1 We have I = # log 2x dx = x - 1, if 1 < x # 2 = # log 2x $ 1dx 2 1 2 Hence # | 1 - x | dx 0 = #0 (1 - x) dx + # (x - 1) dx 1 = log 2x $ # dx - # ' 1 $ 2 # dx 1 dx 2x 2 2 1 2 = bx - x l + b x - x l 1 2 0 2 = x log 2x - # $ xdx 1 x = ;b1 - l - 0E + ;b - 2 l - b 1 - 1lE 1 4 = x log 2x - x + c 2 2 2 =1 Thus (c) is correct option. Thus (c) is correct option. CHAPTER 7 Integrals Page 247 (c) (A) is true but (R) is false. = x (log x - 1) + k (d) (A) is false but (R) is true Formula of Integration by parts: Sol : OD 2010 # (uv) dx =u # vdx - # &dxd (u) # vdx 0 dx Therefore both (A) and (R) are true and (R) is the p We have I = # - p 2 sin2 x dx correct explanation of (A). 2 Let f (x) = sin2 x Thus (a) is correct option. f (- x) = sin2 (- x) p 46. Assertion (A) : # -p 2 sin7 x dx = 0 = sin2 x 2 Reason (R) : Here sin7 x is odd function. = f (x) (a) Both (A) and (R) are true and (R) is the correct Hence, f (x) is an even function and for even function explanation of (A). we have a a (b) Both (A) and (R) are true but (R) is not the # -a f (x) dx = 2 # 0 f (x) dx correct explanation of (A). p (c) (A) is true but (R) is false. Therefore, I =2 # 0 2 sin2 xdx (d) (A) is false but (R) is true 1 - cos 2x dx p Sol : Delhi 2010 =2 #2 0 2 p # 2 p =x 0- sin 2 xp 2 2 We have I = sin7 x dx 2 0 -p 2 = a p - 0k - 1 (sin p - sin 0) Let, f (x) = sin7 x 2 2 = p Here f (- x) = sin7 (- x) 2 Therefore both (A) and (R) are true and (R) is the = (- sin x) 7 correct explanation of (A). = - sin7 x Thus (a) is correct option. = - f (x) 45. Assertion (A) : # log xdx = x (log x - 1) + k Hence, f (x) is odd function and we have p Reason (R) : d # 2 sin7 x dx = 0 # (uv) dx = u # # & dx (u) vdx 0 dx. vdx - # -p 2 Both (A) and (R) are true and (R) is the correct (a) Both (A) and (R) are true and (R) is the correct explanation of (A). explanation of (A). Thus (a) is correct option. (b) Both (A) and (R) are true but (R) is not the p correct explanation of (A). 47. Assertion (A) : # - p2 2 ` sin x + cos x j dx = 2 (c) (A) is true but (R) is false. a a (d) (A) is false but (R) is true Reason (R) : # -a f (x) dx = 2 #0 f (x) dx , Sol : SQP 2020, Foreign 2017 where f (x) is an even function. (a) Both (A) and (R) are true and (R) is the correct We have I = x (log x - 1) + k explanation of (A). (b) Both (A) and (R) are true but (R) is not the = # 1 log xdx II I correct explanation of (A). Using integration by parts, we get (c) (A) is true but (R) is false. (d) (A) is false but (R) is true I = log x $ # dx - # b dxd log x # 1dx ldx Sol : Comp 2009 = log x $ x - # b 1 # x l dx p x = x $ log x - x + k Let, I = # -p 2 ` sin x + cos x j dx 2 CHAPTER 7 Integrals Page 249 (a) Both (A) and (R) are true and (R) is the correct xn = tan2 q explanation of (A). nxn - 1 dx = 2 tan q sec2 qdq (b) Both (A) and (R) are true but (R) is not the p/2 correct explanation of (A). 3 dx = 2 (c) (A) is true but (R) is false. So, # 1 + xn n # tan 1 - 2 + 2/n dq 0 0 (d) (A) is false but (R) is true p/2 Sol : =2 # tan (2/n) - 1 qdq OD 2012 n 0 As We know that In RHS, substituting xn = sin2 q we have b c b # f (x) dx = # f (x) dx + # f (x) dx, Where c e nxn - 1 dx = 2 sin q cos q dq a a c (a.b) 1 p/2 dx =2 So, given reason is true. So, # (1 - xn) 1/n n # cos1 2/n q $ sin2/n - 1 q cos q dq 0 0 p/2 p/2 Now, # sin x dx sin x, =2 # tan (2/n - 1) q dq - p/2 n 0 Since, sin x is an even faction Hence, assertion is true but reason is false. p/2 p/2 Thus (c) is correct option. so, # sin x dx = 2 # sin x dx 2p - p/2 0 p/2 53. Assertion (A) : # sin3 x dx = 0 0 =2 # sin x dx Reason (R) : sin3 x is an odd function O p/2 (a) Both (A) and (R) are true and (R) is the correct =2 # sin x dx explanation of (A). O (b) Both (A) and (R) are true but (R) is not the = - (cos x) 0p/2 correct explanation of (A). - 2 (0 - 1) = 2 (c) (A) is true but (R) is false. Hence, both Assertion and reason are true bat reason (d) (A) is false but (R) is true is not correct explanation for assertion. Sol : SQP 2016 Thus (b) is correct option. 2p I = #0 sin3 x dx 52. If n 2 1, then 2p 1 Assertion (A) : # ! dx = # dx = # (1 - cos2 x) sin x dx 1 + xn (1 - xn) 1/n 0 0 0 Substituting cos x = t we have b b Reason (R) : # f (x) dx = # f (a + b + x) dx sin x dx = - dt a a 1 (a) Both (A) and (R) are true and (R) is the correct explanation of (A). Then I = # (1 - t2) (- dt) = 0 1 (b) Both (A) and (R) are true but (R) is not the Thus assertion is true, Reason is true; Reason is not a correct explanation of (A). correct explanation for Assertion. (c) (A) is true but (R) is false. Thus (b) is correct option. (d) (A) is false but (R) is true 54. Assertion (A) : Sol : # sin 3x cos 5x dx = - cos 8x + cos 2x + C Comp 2015, OD 2011 16 4 As we know that, Reason (R) : 2 cos A sin B = sin (A + B) - sin (A - B) b b (a) Assertion is true, Reason is true; Reason is a # f (x) dx = # f (a + b - x) dx correct explanation for Assertion. a a So, given Reason is false. (b) Assertion is true, Reason is true; Reason is not a Now, in L.H.S, put correct explanation for Assertion.  CBSE Exam Feb 2025 Questions Bank Class 12th Mathematics Click and download any file. All files link are working. Do not assume that links are not working. Open in Google Drive PDF Viewer Only. Don’t use WPS. 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App. AI English Comm. िहदंी अ िहदंी ब संसक ् तृ Class 8th Maths Science Social Science English िहदंी संस्कतृ Class 7th Maths Science Social Science English िहदंी संस्कतृ Class 6th Maths Science Social Science English िहदंी संस्कतृ  Class 5th Maths English िहदंी EVS Class 4th Maths English िहदंी EVS Class 3th Maths English िहदंी EVS Class 2th Maths English िहदंी EVS Class 1th Maths English िहदंी EVS Prep/KG 3rd Year Maths English िहदंी EVS Prep/KG 2nd Year Maths English िहदंी EVS Prep/KG 1st Year Maths English िहदंी EVS CHAPTER 7 Integrals Page 251 59. Find # sin2 x - cos2 x dx. = # sec2 xdx sin2 x cos2 x Sol : Foreign 2014 = tan x + C sin2 x - cos2 x dx Write the value of # dx. We have I = # 64. x2 + 16 sin2 x cos2 x sin2 x cos2 x dx Sol : Foreign 2011 = # 2 2 dx - # sin x cos x sin2 x cos2 x dx = # sec x dx - # cosec x dx 2 2 We have I = # x2 + 16 = tan x + cot x + C = # 2 dx 2 # x2dx = 1 tan-1 x + C x + (4) + a2 a a 60. Find # sin6 x dx. = tan-1 x + C 1 4 4 cos8 x Sol : Foreign 2014 2 - 3 sin x dx. 65. Write the value of # cos2 x 6 sin x dx = # tan6 x sec2 x dx Sol : Delhi 2011 We have I = # cos8 x Substituting tan x = t & sec2 x dx = dt we have We have I = 2 - 3 sin x dx # cos2 x I = # t6 dt = # b 22 - 3 sin2 x l dx 7 cos x cos x = t +C = # (2 sec2 x - 3 sec x tan x) dx 7 7 = tan x + C = 2 # sec2 x dx - 3 # sec x tan x dx 7 = 2 tan x - 3 sec x + C 61. Evaluate # cos-1 (sin x) dx. Sol : Foreign 2014 66. Write the value of # sec x (sec x + tan x) dx. Sol : Delhi 2011 We have I = # cos-1 (sin x) dx = # cos-1 9cos a p2 - x kC dx We have I = # sec x (sec x + tan x) dx = # (sec2 x + sec x tan x) dx = # a p2 - x k dx cos-1 (cos q) = q = # sec2 x dx + # sec x tan x dx = p # dx - # x dx 2 = tan x + sec x + C 2 = px- x +C 2 2 dx. 67. Evaluate # 62. Evaluate # (1 - x) x dx. 1 - x2 Sol : OD 2011, Delhi 2008 Sol : Delhi 2012, Comp 2010 We have I = # dx We have I = # (1 - x) x dx 1 - x2 = # dx = #( x - x x ) dx (1) 2 - x2 = # (x1/2 - x3/2) dx = sin-1 x + C # dx = sin-1 x + C a a2 - x2 = 2 x3/2 - 2 x5/2 + C -1 3 5 68. Evaluate # e tan x dx. 1 + x2 63. Evaluate 2 # 1 + cos dx. Sol : OD 2011 2x tan-1 x Sol : Foreign 2012 We have I =e # dx 1 + x2 2 We have I = # 1 + cos 2x dx Substituting tan-1 x = t & 1 2 dx = dt we have 1+x = # 2 dx tan x -1 cos 2q = 2 cos2 q - 1 I = # e dx = # et dt 2 cos2 x 1 + x2 CHAPTER 7 Integrals Page 253 = # sec2 x dx Substituting x2 = t & 2x dx = dt & dx = dt 2x = tan x + C When x = 0 , then t = 0 and when x = 1, then t = 1, x et dt 1 dx Thus I = #0 2x 77. Find #. x2 + 4x + 8 Sol : = 1 # e tdt 1 Delhi 2017 2 0 dx 1 We have I = # = [et] 10 2 x2 + 4x + 8 = # dx = 1 [e1 - e 0] x2 + 4x + 4 + 4 2 1 = [ e - 1] = # dx 2 (x + 2) 2 + (2) 2 p/4 82. Evaluate #0 sin 2x dx. = 1 tan-1 b x + 2 l + C 2 2 Sol : Foreign 2014 78. Find : # 3 - 5 sin x dx. p/4 cos2 x We have I = #0 sin 2x dx Sol : Foreign 2018 = :- cos 2x D p/4 3 - 5 sin x dx 2 0 We have I = # cos2 x = - 1 [cos 2x] r0 /4 2 = # b 32 - 5 sin2 x l dx cos x cos x = - 1 9cos 2 p - cos 0C 2 4 = 3 # sec x dx - 5 # sec x tan x dx 2 = - 1 9cos p - 1C = 3 tan x - 5 sec x + C 2 2 = - [0 - 1] = 1 1 3 2 2 79. Evaluate #2 3x dx. 83. Evaluate #0 1 1 dx. Sol : Delhi 2017 Sol : 1 - x2 Comp 2014, OD 2011 x 3 3 dx = c 3 m 3 #2 x log 3 2 1 1 We have I = #0 dx 1 - x2 = 1 [3 x] 23 log 3 = [sin-1 x] 10 # 1 dx = sin-1 x + C 1 - x2 = 1 - 2 log 3 [3 2 ] = sin-1 1 - sin-1 0 3 = 1 (27 - 9) = sin-1 a sin p k - sin-1 (sin 0) log 3 2 = 18 p = -0 = p log 3 2 2 p/4 80. Evaluate #0 tan x dx. 84. Evaluate # 2 x3 - 1 dx. Sol : Foreign 2014, Delhi 2013 1 x2 Sol : Comp 2014 p/4 #0 tan x dx = [log sec x ] p0 /4 x3 - 1 dx 2 We have I = # 1 x2 = log sec p - log sec 0 = # ax - 12 k dx 2 4 1 x = log 2 - log 1 2 2 = :x + 1 D 2 x 1 = 1 log 2 (2) 2 1 (1) 2 1 =c + m-c + m 2 1 2 2 2 1 #0 2 81. Evaluate x ex dx. Sol : = b 2 + 1 l - b 1 + 1l = 1 Foreign 2014 2 2 1 #0 2 We have I = x ex dx CHAPTER 7 Integrals Page 255 = [log 2 - log 1] Comparing coefficient of x2 we have = log 2 - 0 A + C = 0 & A = - C = - 34 = log 2 Thus given integral becomes I= # - 3 dx + 1 # dx + 3 # dx 4 (x + 1) 2 (x + 1) 2 4 (x - 1) = - 3 log (x + 1) + 1 ; - 1 E + 3 log (x - 1) + C 4 2 (x + 1) 4 SHORT ANSWER QUESTIONS = - 3 log (x + 1) - 1 + 3 log (x - 1) + C 4 2 (x + 1) 4 e cos x = 3 log b x - 1 l - 1 p 93. Evaluate : #0 e cos x + e- cos x dx 4 x+1 2 (x + 1) +C Sol : OD 2024 log 3 1 Let I = #0 p e cos x dx...(1) 95. Evaluate #log ^ex + e-x h^ex - e-x h dx e cos x + e- cos x 2 a a Sol : OD 2023 Using #0 f (x) dx = #0 f (a - x) dx we have log 3 1 I = e cos (p - x) #0 p dx Let I = #log 2 ^ex + e-x h^ex - e-x h dx cos (p - x) e + e- cos (p - x) log 3 1 I = # - cosex p - cos x dx...(2) I = #log ^e2x - e-2x h dx 0 e + e cos x 2 Adding equations (1) and (2), we get e2x dx log 3 = #log 2 ^e - 1h 4x 2I = #0 e cos x p dx + #0 p e- cos x dx e + e- cos x cos x e - cos x + e cos x Substituting t = e2x & dt = 2e2x dx e2`log 3j p cos x - cos x = # e cos x + e- cos x dx = 3 and e2`log 2j =2 0 e +e 3 1 = #0