Chapter 7.1 Overview ∫ x dx7.1.1 Let d F (x) = f (x). Then, we write f () =F (x) + C. These integrals aredx called indefinite integrals or general integrals, C is called a constant of integration. All these integrals differ by a constant. 7.1.2 If two functions differ by a constant, they have the same derivative. ∫ x dx7.1.3 Geometrically, the statement f () =F (x) + C = y (say) represents a family of curves. The different values of C correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. Further, the tangents to the curves at the points of intersection of a line x = a with the curves are parallel. 7.1.4 Some properties of indefinite integrals (i) The process of differentiation and integration are inverse of each other, x dx = () f '() fx +i.e., df () fx and x dx = () C , where C is anydx ∫ ∫ arbitrary constant. (ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. So if f and g are two functions such that d ∫ d ∫ ∫ x()x dx = () f () and dxf gxdx , then ∫ xdxg ()are equivalent.dx dx (iii) The integral of the sum of two functions equals the sum of the integrals of f ()+g ())f () g ()dxthe functions i.e., ∫( x xdx = ∫ x dx + ∫ x . (iv) A constant factor may be written either before or after the integral sign, i.e., af ()dx af ()dx ∫ x = ∫ x , where ‘a’ is a constant. (v) Properties (iii) and (iv) can be generalised to a finite number of functions f1, f, ..., f and the real numbers, k, k2, ..., kgiving2n 1n kf ()+kf () ... , kf () dx k f ()dx +kf ()dx ... ++ ++ kf ()dx∫(11 x 22 x nnx ) = 1 ∫1 x 2 ∫2 xn ∫nx 7.1.5 Methods of integration There are some methods or techniques for finding the integral where we can not directly select the antiderivative of function f by reducing them into standard forms. Some of these methods are based on 1. Integration by substitution 2. Integration using partial fractions 3. Integration by parts. 7.1.6 Definite integral b ∫ x dxThe definite integral is denoted by f () , where a is the lower limit of the integral a and b is the upper limit of the integral. The definite integral is evaluated in the following two ways: (i) The definite integral as the limit of the sum b (ii) f ()dx = F(b) – F(a), if F is an antiderivative of f (x).∫ x a 7.1.7 The definite integral as the limit of the sum b The definite integral f ()dx is the area bounded by the curve y = f (x), the ordi∫ x a nates x = a, x = b and the x-axis and given by b 1f () = (b – a) lim ⎡fa +f ( +)+... f (a +(n –1)h)⎤∫ x dx () ah n→∞n ⎣ ⎦ a or b f () = lim h ⎡fa() +f (ah)+ + ... f (+( ⎦,∫ x dx ⎣+ an –1)h)⎤ h→0 a –bawhere h = →0 as n →∞. n 7.1.8 Fundamental Theorem of Calculus (i) Area function : The function A (x) denotes the area function and is given x by A (x) = ∫f ()x dx . a (ii) First Fundamental Theorem of integral Calculus Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function . Then A′ (x) = f (x) for all x ∈ [a, b] . (iii) Second Fundamental Theorem of Integral Calculus Let f be continuous function defined on the closed interval [a, b] and F be an antiderivative of f. b ∫f ()x dx = [()]ba = F(b) – F(a).F x a 7.1.9 Some properties of Definite Integrals bb P : ∫f xdx f ()() ∫ t dt= 0aa ba a ∫ x dx () f ()P1 : f () = – ∫f xdx , in particular, ∫ x dx = 0 ab a b cb f () f ()dx + f (): ∫ x dx = ∫ x ∫ x dx P2a ac b b P3 : f () = ∫f (ab+)dx∫ x dx– x a a a a P : ∫f ()x dx = ∫f (ax)dx– 400 2a aa f ()dx f ()dx + f ()dx P5 : ∫ x = ∫ x ∫ 2–ax 0 00 2a ⎧a P : ∫f ()x dx = 2 f ()dx,if f (2 a −=x) f () x ,6 ⎪∫ x 0 ⎨0 ⎪⎩0, if f (2 ax) −f ( ). −=x a a f () ∫ x dxP7 : (i) ∫ x dx = 2 f () , if f is an even function i.e., f (–x) = f (x) – a 0 a ∫ x dx(ii) f () = 0, if f is an odd function i.e., f (–x) = –f (x) – a 7.2 Solved Examples Short Answer (S.A.) ⎛2ab 32 ⎞ –2 +3c x ⎟ w.r.t. xExample 1 Integrate ⎜⎝ xx ⎠⎛2ab 32 ⎞ –2 +3c x dxSolution ∫⎜ ⎟⎝ xx ⎠ 2–1 = ∫2a ()2 dx – bx –2 dx +∫33x ∫ cx dx 5 b 9cx 3 = 4ax + + +C . x 5 3axExample 2 Evaluate dx2 22b cx Solution Let v = b2 + c2x2 , then dv = 2c2 xdx 3ax 3adv Therefore, ∫2 22 dx = b +cx 2c2 v3a 2 22b cx C.= 2 log 2c Example 3 Verify the following using the concept of integration as an antiderivative. 3 23xdx xx x – – log x 1C x 1 23 dx2 x3 Solution x – – log x 1Cdx 23 2x 3x21= 1 – – 23 x 1 1 x3 = 1 – x + x2 – = . x1 x 1 23 3⎛ xx ⎞ xThus ⎜ x – + – log x +1 + C⎟ =∫ dx 23 x +1⎝ ⎠ 1 x Example 4 Evaluate 1– x dx , x ≠ 1. 1 x dx1+ x –1Solution Let I= dx = ∫2 dx + 2 = sin x + I1 ,∫1– x 1– x 1– xxdx where I1 = Put 1 – x2 = t2 ⇒ –2x dx = 2t dt. Therefore I1 = – dt = – t + C = –1– x2C Hence I = sin–1x –1– x2 C. dx , β >αExample 5 Evaluate ∫(x )– x)– αβ( 2 22Solution Put x – α = t2. Then – x = – t= – t – = – t – and dx = 2tdt. Now 2tdt 2dtI= = ∫β ––t2 )∫ t2 (βα–– t2 )(α 22 dt 2 , where k2– k – t –1 t –1 x – α2sin +=C 2sin +C= .k –βα Example 6 Evaluate ∫tan 8 xsec 4 xdx Solution I = ∫tan 8 xsec 4 xdx 82 2= tan x sec x sec x dx ∫() 82 2= ∫tan x(tan x +1)sec x dx 102 82 = tan x sec xdx +tan x sec xdx ∫∫ 11 9tan x tan x = ++C .11 9 3xExample 7 Find ∫ dx x4 +3x2 +2 Solution Put x2 = t. Then 2x dx = dt. x3dx 1 t dt Now I = ∫42 =∫2x +3x +22 t 3t 2++ t AB Consider 2 =+ t ++3t 2 t +1 t +2 Comparing coefficient, we get A = –1, B = 2. 1 ⎡ dt dt ⎤Then I = ⎢2∫ – ∫⎥2 ⎣ t +2 t +1 ⎦ 1 = ⎡2log t 2+−log t +1 ⎤⎣ ⎦2 x2 +2log +C= Example 8 Find ∫ 2 dx 22sin x +5cos x Solution Dividing numerator and denominator by cos2x, we have sec 2 x dx I = 22tan x 5 Put tanx = t so that sec2x dx = dt. Then dt 1 dt = 2I = ∫2 ∫2t +52 2 ⎛5 ⎞ t +⎜ ⎟2⎝ ⎠ 12 –1 ⎛2t ⎞ = tan +C⎜ ⎟ 25 5⎝ ⎠ 1 –1 ⎛2 tan x ⎞ = tan +C .⎜ ⎟ 10 5⎝ ⎠ 2 Example 9 Evaluate 7– 5x dx as a limit of sums. –1 21+Solution Here a = –1 , b = 2, and h = , i.e, nh = 3 and f (x) = 7x – 5. n Now, we have 2 (7x –5)dx =lim h ⎡f –1 +f (–1 h) f (–1 +2h ... f (–1+( )⎤() ++ )++ n –1 h)∫ h→0 ⎣⎦ –1 Note that f (–1) = –7 – 5 = –12 f (–1 + h) = –7 + 7h – 5 = –12 + 7h f (–1 + (n –1) h) = 7 (n – 1) h – 12. Therefore, 2 7x –5 dx =limh⎡–12 +(7h – 12) + h ... (7 n –1 h –12)⎤.∫() ⎣() (14 –12) ++ () ⎦h→0 –1 = lim hh71 2... (n –1)⎤– 12n⎤⎡ ⎡+++⎣ ⎦⎦h→0 ⎣ ⎡(n –1)n ⎤⎡7 ⎤lim h 7h – .12 n nh nh – )–12 nh = ⎢⎥= lim ()( h h→02 h→0 ⎢2 ⎥⎣⎦⎣ ⎦ 7 79× –9 = 33–0 –12 3 = –36 = .222 π 27tan xExample 10 Evaluate dx∫ 77cot x +tan x0 Solution We have π 27tan xI = dx ...(1)∫ 77cot x +tan x0 π 7 ⎛π ⎞tan – x2 ⎜⎟⎝2 ⎠ = ∫ dx by (P)7 ⎛π ⎞ 7 ⎛π ⎞ 40 cot – x +tan – x⎜⎟ ⎜⎟⎝2 ⎠⎝2 ⎠ π cot xdx 27 ()= ...(2)∫ 77 0 cot x dx +tan x Adding (1) and (2), we get π 27 7⎛tan x +cot x ⎞2I =⎜ dx∫ 77 ⎟tan x +cot x0 ⎝⎠ π 2 0 dx=∫ which gives I π 4 . Example 11 Find 8 2 ∫ 10 – 10 – x dx x x+ Solution We have I = 8 2 ∫ 10 – 10 – x dx x x+ ...(1) = 8 2 10 – (10 – ) 10 – 10 – 10 – x x x dx by (P3) ⇒ 8 2 I= ∫ 10 – x x + dx x (2) Adding (1) and (2), we get 2I 8 2 1dx 8 – 2 6 Hence I = 3 Example 12 Find 4 0 π ∫ Solution We have I = 4 0 π ∫ 1sin 2 x dx+ = 4 0 π ∫ ( )2sin cos x x dx+ = ( ) 4 0 sin cos x xdx π +∫ π = (−cos x+sin x)4 0 I = 1. Example 13 Find x2tan–1xdx. Solution I = x2tan –1 xdx3 –12 1 x = tanx xdx–. dx∫∫21 +x 3 x3 –11 ⎛ x ⎞ tan x– ⎜x−⎟dx= ∫23 31 +x⎝⎠ x3 –1 x21 = tan x– + log 1 +x2 +C.3 66 Example 14 Find ∫ Solution We have 2 22I = 10 –4 x 4xdx = 2x–1 3 dx Put t= 2x– 1, then dt= 2dx. 12 2Therefore, I=t +()dt 3 2 ∫ 1 t29 9 = t log tt29 C22 4 1 29 =(2x–1)( 2x–1) 9 log ++ (2x–1)+ (2x–1)2+9 +C.4 4 Long Answer (L.A.) x2 dx Example 15 Evaluate ∫ 42 . x + x − 2 Solution Let x2 = t. Then x2 tt AB == =+42 2x + x −2 +− 2 (2)( t + t −1) t + 2 t −1 So t = A (t – 1) + B (t + 2) tt Comparing coefficients, we get A = 2, B = 1.33 x2 21 11 =+So42 2 2x + x −23 x + 23 x −1 Therefore, x2 211 dxdx = dx +∫ 42 ∫ 2 ∫ 2x + x − 23 x + 23 x −1 21 x 1 x−1tan –1 + log +C= x +132 26 3xxExample16 Evaluate dx x4–9 Solution We have 33xx xxdx I = dx = dx = I1+ I2 .4 44x –9 x –9 x–9 3xI =Now 1 ∫ 4x –9 Put t = x4 – 9 so that 4x3 dx = dt. Therefore 1 dt 1 1 1 = log I = tC1 = log x4–9 + C14 t 4 4 x dxI =Again, 2 4.x –9 Put x2 = u so that 2x dx = du. Then 1 du 1 u –3 = 2 2= log C2I2 2 u –3 26 u 3 1 x2 –3 = log + C2 .12 x2 + 3 Thus I = I1 + I2 1 1 x2 –3 = log x4 –9 + log +C .4 12 x2 +3 π 2 sin 2 x 1Example 17 Show that = log( 2 +1) ∫ 0 sin x + cos x 2 Solution We haveπ 2 sin2 xI = dx∫ 0 sin x + cos x = 2 2 0 πsin – 2 π πsin cos – – 2 2 x dx x x π ⎛ ⎞⎜ ⎟⎝ ⎠ ⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ∫ (by P4) ⇒ I = π 22 0 cos sin cos x dx x x+∫ Thus, we get 2I = π 2 0 1 π2 cos – 4 dx x ⎛ ⎞⎜ ⎟⎝ ⎠ ∫ = π 2 0 1 π sec – 42 dxx ⎛ ⎞⎜ ⎟⎝ ⎠∫ = 2 0 1 π πlog sec tan – – 4 42 x x ⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ π = 1 π πlog sec tan 4 42 ⎡ ⎛ +⎜⎢ ⎝⎣ π π– logsec tan – 4 4 ⎤⎞ ⎛ ⎞ ⎛ ⎞+ −⎟ ⎜ ⎟ ⎜ ⎟⎥⎠ ⎝ ⎠ ⎝ ⎠⎦ = (1 log 2 ⎡⎣ ) (– log 21+ )21 ⎤− ⎦ = 1 log 2 21 2–1 + = (1 log 2 ⎛ ⎜ ⎝ )2 21 1 ⎞+ ⎟ ⎠ = (2 log 2 )21+ Hence I = (1 log 2 )21+ . Example 18 Find ( ) 1 2–1 0 tan x dxx∫ 1 –1Solution I = ∫ x(tan x)2 dx . 0 Integrating by parts, we have x2 21 11 2 tan –1 x⎡ –1 ⎤ x .2 dxI = (tan x) – ∫22 ⎣⎦020 1+ x 2 1 2π x –1 = – ∫2 .tan x dx 32 01+ x tan –1 xdx= π2 – I1 , where I1 = ∫ 1 x2232 01+ x 1 x2 +1–1 Now I1 = ∫2 tan–1x dx0 1+ x 11 –11 –1 = ∫tan x dx – ∫2 tan x dx 0 01+ x = I2 – 1 ((tan –1 x)2 )1 = I2 – π2 2 032 1 1 –1Here I2 = ∫tan –1 x dx = ( x tan x)10 – ∫x 2 dx 0 01+ x π 1 π 1=– (log 1 + x2 )10 = –log2.42 42 π 1 π2 Thus I1 = –log2−42 32 2 2 π2ππ 1 ππ 1Therefore, I = – log2 = – + log 2 3242 321642 π2 –4π = +log 16 2 Example 19 Evaluate ∫f ()xdx , where f (x) = |x + 1| + |x| + |x – 1|. –1 ⎧2– x, if –1 <≤x 0 ⎪Solution We can redefine f as f ()= x +2,if 0 x 1x ⎨ <≤ ⎪⎩3,x if 1 x 2<≤ 20 12 ()dx =( )dx +(+2)dx +3x dx Therefore, ∫fx ∫2– x ∫x ∫ (by P2)–1–10 1 0 1222 2⎛ x ⎞⎛x ⎞⎛3x ⎞++= ⎜2–x ⎟ ⎜+2x⎟⎜ ⎟⎝ 2 ⎠–1 ⎝2 ⎠0 ⎝2 ⎠1 ⎛ 1 ⎞⎛ 1 ⎞⎛41 ⎞ 5 5 919 = 0– ⎜–2 – ⎟⎜ + +2⎟+3⎜ – ⎟ = + + = .⎝ 2 ⎠⎝ 2 ⎠⎝22 ⎠2222 Objective Type Questions Choose the correct answer from the given four options in each of the Examples from 20 to 30. xExample 20 ∫e (cos x – sin x)dx is equal to (A) ex cos x +C (B) ex sin x +C (C) –cos ex x +C ex(D) –sin x +C x ' x⎡ x () = f CSolution (A) is the correct answer since ∫e ⎣f ()+fx ⎤⎦dx e ()x + . Here f (x) = cosx, f′ (x) = – sin x. Example 21 ∫ 2 dx 2 is equal tosin xcos x (A) tanx + cotx + C (B) (tanx + cotx)2 + C (C) tanx – cotx + C (D) (tanx – cotx)2 + C Solution (C) is the correct answer, since 22dx (sin x +cos x)dx I = ∫2 2 = ∫22sin xcos x sin x cos x = ∫sec 2 x dx +∫cosec 2 x dx = tanx – cotx + C x – x3–5 ee Example 22 If ∫ x – x dx = ax + b log |4ex + 5e–x| + C, then4e +5e –17 17(A) a = , b = (B) a = , b = 88 88 –1–7 1–7 (C) a = , b = (D) a = , b = 88 88 Solution (C) is the correct answer, since differentiating both sides, we have x – xx – x4–5 e3–5 ee (e )= a + b ,x – xx – x4e +5e 4e +5egiving 3ex – 5e–x = a (4ex + 5e–x) + b (4ex – 5e–x). Comparing coefficients on both sides, we get 3 = 4a + 4b and –5 = 5a – 5b. This verifies a = –1, b = 7.88 bc ∫+ Example 23 f ()x dx is equal to ac+ bb f ( – ) f (xc)dx(A) ∫ xcdx (B) ∫+ aa b –bc (C) ∫f ()x dx (D) ∫f ()x dx a –ac Solution (B) is the correct answer, since by putting x = t + c, we get bb I = ∫(ct )dt = ∫f (+ .f + xc)dx aa Example 24 If f and g are continuous functions in [0, 1] satisfying f (x) = f (a – x) a and g (x) + g (a – x) = a, then ∫f () () x .g xdx is equal to 0 a ∫ x dx(A) a (B) af ()2 20 aa ∫ x dx ()(C) f () (D) a ∫f xdx 00 a f () () .g dx Solution B is the correct answer. Since I = ∫ xx 0 ∫ a ∫ a = f (ax– )( g – )dx = f ()( x )dxax xag– () 00 aa ∫a af ()dx – x .g = af ()dx – I= ∫ xf () () x dx ∫ x000 aa or I = ∫f ()x dx .20 y dt 2dyExample 25 If x = ∫ 2and 2 = ay, then a is equal to 019+t dx(A) 3 (B) 6 (C) 9 (D) 1 y dt dx 1 =Solution (C) is the correct answer, since x = ∫ 019t2 ⇒ dy +y2+ 19 2 18y dydywhich gives 2 = 2 . = 9y.dx 21+9y dx 1 x3 ++1xExample 26 dx is equal to ∫2 –1 x +2 +1 1 x (A) log 2 (B) 2 log 2 (C) log2 (D) 4 log 22 1 x3 ++1x dxSolution (B) is the correct answer, since I = ∫2 –1 x +2 +1x 13 1x +11 x = ∫2 ∫2 dx = 0 + 2 ∫+12 dxx –1 x +2 +1 –1 x +2 ++10 (+1)x x x [odd function + even function] 1 x +1 11dx =2 dx= 2 ∫ 2 ∫ =2 1 = 2 log 2.log x +1 00 () x +10 x +1 1 t 1 t ∫ e ∫ e 2Example 27 If dt = a, then dt is equal to 01+ t 0 (1+ t ) ee ee(A) a – 1 + (B) a + 1 – (C) a – 1 – (D) a + 1 +22 22 1 te Solution (B) is the correct answer, since I = ∫dt 01+ t 11 t1 t e e= +∫ 2 dt = a (given)1+ t 00 (1+ t ) 1 t Therefore, ∫ e 2= a – e + 1. 0 (1+ t) 2 2 Example 28 dx is equal toxcos πx∫ –2 84 21(A) (B) (C) (D)ππ ππ 22 dx =2 ∫ dxxcos πx x cos πxSolution (A) is the correct answer, since I = ∫–2 ⎧13 22 2 dx + dx +xcos πx x cos πx x cos πx= 2 ⎨⎪∫ ∫ ∫⎪01 3 ⎩ 22 Fill in the blanks in each of the Examples 29 to 32. sin6 xExample 29 ∫ 8 dx = _______. cos x 0 ⎫ dx⎪⎬ ⎪⎭ = 8 π . tan7 xSolution +C 7 a f ()dxExample 30 ∫ x = 0 if f is an _______ function. – a Solution Odd. 2aa ()Example 31 ∫f ()xdx = 2 ∫f xdx , if f (2a – x) = _______. 00 Solution f (x). π 2 sin n xdxExample 32 = _______.∫ nn 0 sin x +cos x πSolution .4 7.3 EXERCISE Short Answer (S.A.) Verify the following : ∫2–1 x1. dx = x – log |(2x + 3)2| + C2x +3 2x +32. ∫2 dx = log |x2 + 3x| + Cx +3x Evaluate the following: 6log x 5log x2(x +2)dx e – e 3. ∫ 4. ∫4log x 3log x dx e – ex +1 5. 7. ( )1cos sin x dx x x + +∫ 2 4tan xsec xdx ∫ 6. 8. 1cos dx x+∫ sin cos 1sin 2 x x dx x + +∫ 9. ∫ 10. ∫ 1 x dx x + (Hint : Put x = z) 11. ∫ – ax ax + 12. 1 2 3 41 x dx + x ∫ (Hint : Put x = z4) 13. ∫ 2 4 1 x dx x + 14. ∫ 216 – 9 dx x 15. ∫ 3– 2 2 dt t t 16. ∫ 2 3– 1 9 x dx x + 17. ∫ 25– 2x + xdx 18. 4 –1 x dx x∫ 19. 2 41– x dx x∫ put x2 = t 20. ∫ 22–ax x dx 21. ( ) –1 3 2 2 sin 1– x dx x ∫ 22. ( )cos5 cos 4 1– 2cos3 x x dx x +∫ 23. 6 6 2 2 sin cos sin cos x x dx x x +∫ x cos x – cos2 x24. ∫ dx 25. ∫ dx a3– x3 1–cos x dx 26. ∫ (Hint : Put x2 = sec θ)xx4 –1 Evaluate the following as limit of sums: 2 2 x27. ∫( x2 + 3)dx 28. ∫edx 0 0 Evaluate the following: π 1 dx 2 tan x dx29. ∫ x – x 30. 22 0 e + e ∫ 01+ m tan x 2 dx 1 xdx 31. ∫ 32. ∫ 2 1 x –1 (2 − x)() 01+ x 1 2π 2 dx 33. ∫ xsin x cos xdx 34. ∫ 0 0(1+ x2) 1− x2 (Hint: let x = sinθ) Long Answer (L.A.) 22xdx xdx 35. ∫ 42 36. 2 22212 (xa )( xb )x – x –π x 2–1 x 37. ∫ 38. ∫ dx + ( x –10 1sin x )( x + 2)( x –3 ) tan –1 x ⎛1++xx2 ⎞39. ∫e ⎜ 2 ⎟dx 40. ∫sin–1 1+x⎝⎠ (Hint: Put x = a tan2θ) −3x 341. 5 42. ∫e cos x dx π 3 (1−cos x)2 43. ∫tan x dx (Hint: Put tanx = t2) π 2 dx44. ∫2 2 222 0 (cos x +b sin a x) (Hint: Divide Numerator and Denominator by cos4x) 1 π x log(1 2 ) xdx +xdx xlog sin 45. ∫ 46. ∫ 0 0 π 4 log (sin x +cos xdx )47. ∫ π− 4 Objective Type Questions Choose the correct option from given four options in each of the Exercises from 48 to 63. cos2 x – cos2θ48. dx is equal to∫cos x – cosθ (A)2(sinx + xcosθ) + C (B) 2(sinx – xcosθ) + C (C)2(sinx + 2xcosθ) + C (D) 2(sinx – 2x cosθ) + C dx49. is equal tosin x – a sin x – bsin(x – b) sin(x – a)(A) sin (b – a) log + C (B) cosec (b – a) log + C sin(x – a) sin(x – b) sin(x – b) sin(x – a)(C) cosec (b – a) log + C (D) sin (b – a) log + C sin(x – a) sin(x – b) 50. ∫tan–1 x dx is equal to (A) (x + 1) tan–1 x – x + C (B) x tan–1 x – x + C (C) xx–tan–1 x + C (D) x – ( x +1)tan –1 x + C x ⎛ 1– x ⎞2 51. e dx is equal to∫⎜ 2 ⎟⎝1+ x ⎠ ex –ex (A) 2 +C (B) 2 +C1+ x 1+ x ex –ex + C + C(C) ()2 (D) ()21+ x21+ x2 9x52. ∫ 6 dx is equal to2(4x +1) –5 –51 ⎛ 1 ⎞ 1 ⎛ 1 ⎞(A) ⎜ 4 + 2 ⎟+ C (B) ⎜ 4 + 2 ⎟+ C 5x ⎝ x ⎠ 5 ⎝ x ⎠ 1 )–5 1 ⎛ 1 ⎞–5 (C) (14 + C (D) ⎜+ 4 + C+ 2 ⎟10x 10 ⎝ x ⎠ dx 153. If ∫ 2= a log |1 + x2| + b tan–1x + log |x + 2| + C, then( x + 2)( x +1) 5 –1–2 12(A) a = 10, b = (B) a = 10, b = –55 –12 12(C) a = 10, b = (D) a = 10, b = 5 5 3x54. ∫ is equal tox +1 23 23xx xx(A) x ++ –log 1– x + C (B) x + – –log 1– x + C 23 23 23 23xx xx(C) x – – –log 1+ x + C (D) x – + – log 1+ x + C 23 23 x + sin x55. dx is equal to1cos x∫+ (A) log 1cos x + C (B) log + C+ x + sin x x x(C) x –tan + C (D) x.tan + C 2 2 33xdx 22 256. If a(1 x ) b 1 x C, then 1 x2 1 –1(A) a = b = 1 (B) a = , b = 13, 3 –1 1(C) a = , b = –1 (D) a = b = –13,3 π 4dx57. is equal to∫1+ cos2 x–π 4 (A) 1 (B) 2 (C) 3 (D) 4 π π 58. 2 0 ∫ is equal to (A) 22 (B) 2 ( 21)+ (C) 2 (D) (2 )2–1 ∫ 2 sin x59. cos xedx is equal to _______. 0 x +3 x60. ∫edx = ________.(x +4)2 Fill in the blanks in each of the following Exercise 60 to 63. a 1 61. If ∫2 dxπ = , then a = ________.14+ x 80 sin x 62. ∫+ dx = ________.34cos 2 x π 63. The value of ∫ sin3x cos2x dx is _______. −π

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