{"id":37312,"date":"2022-12-01T17:49:28","date_gmt":"2022-12-01T12:19:28","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=37312"},"modified":"2022-12-03T14:15:17","modified_gmt":"2022-12-03T08:45:17","slug":"maths-2b-integration-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2b-integration-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type"},"content":{"rendered":"
Students must practice these Maths 2B Important Questions<\/a> TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. Question 2. Question 3. <\/p>\n Question 4. Question 5. Question 6. Question 7. Question 8. <\/p>\n Question 9. Question 10. Question 11. Question 12. <\/p>\n Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. <\/p>\n Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. <\/p>\n Question 25. Question 26. Question 27. Question 28. Question 29. Question 30. <\/p>\n Question 31. Question 32. Question 33. Question 34. Question 35. Question 36. <\/p>\n Question 37. Question 38. Question 39. Question 40. Question 41. <\/p>\n Question 42. Question 43. Question 44. Question 45. <\/p>\n Question 46. Question 47. Question 48. <\/p>\n Question 49. Question 50. Question 51. Question 52. <\/p>\n Question 53. Question 54. Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type Question 1. Evaluate . [Mar. ’05] Solution: Question 2. Evaluate . [Mar. ’12, ’11, ’10] Solution: … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2B Integration Important Questions Long Answer Type<\/h2>\n
\nEvaluate \\(\\int \\frac{d x}{4+5 \\sin x}\\). [Mar. ’05]
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{5+4 \\cos x}\\). [Mar. ’12, ’11, ’10]
\nSolution:
\nWrite tan \\(\\frac{x}{2}\\) = t
\nDifferentiating on both sides with respect to ‘x’.
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{1+\\sin x+\\cos x} d x\\). [(AP) Mar. ’20; (TS) ’15]
\nSolution:
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{4 \\cos x+3 \\sin x}\\). [(TS) Mar. ’18]
\nSolution:
\n
\n
\n<\/p>\n
\nFind \\(\\int \\frac{d x}{3 \\cos x+4 \\sin x+6}\\). [May ’15 (AP); Mar. ’13]
\nSolution:
\nLet tan \\(\\frac{x}{2}\\) = t
\nDifferentiating on both sides with respect to ‘x’
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{2-3 \\cos 2 x} d x\\). [May ’10, ’05]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{5+4 \\cos 2 x}\\). [May ’13, Mar. ’11]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{\\cos x+3 \\sin x+7}{\\cos x+\\sin x+1} d x\\). [(AP) Mar. ’19; May ’06]
\nSolution:
\nLet cos x + 3 sin x + 7 = \\(\\frac{d}{dx}\\) (cos x + sin x + 1) + \u00b5 (cos x + sin x + 1) + \u03b3
\n= \u03bb(-sin x + cos x) + \u00b5 (cos x + sin x + 1) + \u03b3 …….(1)
\n= -\u03bb sin x + \u03bb cos x + \u00b5 cos x + \u00b5 sin x + \u00b5 + \u03b3
\nComparing the coefficients of cos x on both sides, we get
\n\u03bb + \u00b5 = 1 ……..(2)
\nComparing the coefficients of sinx on both sides, we get
\n-\u03bb + \u00b5 = 3 ………(3)
\nSolving (2) and (3)
\n
\nComparing constant terms on both sides, we get
\n\u00b5 + \u03b3 = 7
\n\u21d2 2 + \u03b3 = 7
\n\u21d2 \u03b3 = 5
\n
\n
\n
\nwhere k is an integration constant.<\/p>\n
\nEvaluate \\(\\int \\frac{2 \\sin x+3 \\cos x+4}{3 \\sin x+4 \\cos x+5} d x\\). [Mar. ’16 (AP & TS); Mar. ’14, ’11]
\nSolution:
\n2 sin x + 3 cos x + 4 = \u03bb(3 sin x + 4 cos x + 5) + \u00b5(3 sin x + 4 cos x + 5) + \u03b3
\n= \u03bb(3 cos x – 4 sin x) + \u00b5(3 sin x + 4 cos x + 5) + \u03b3 …….(1)
\nComparing,
\nCoeff. of cos x, 3 = 3\u03bb + 4\u00b5
\n\u21d2 3\u03bb + 4\u00b5 – 3 = 0 ……….(2)
\nCoeff. of sin x, 2 = -4\u03bb + 3\u00b5
\n\u21d2 4\u03bb – 3\u00b5 + 2 = 0 ……..(3)
\nSolving (2) and (3),
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{9 \\cos x-\\sin x}{4 \\sin x+5 \\cos x} d x\\). [Mar. ’17 (TS), Mar. ’08]
\nSolution:
\n9 cos x – sin x = \u03bb(4 sin x + 5 cos x) + \u00b5(4 sin x + 5 cos x) + \u03b3
\n= \u03bb(4 cos x – 5 sin x) + \u00b5(4 sin x + 5 cos x) + \u03b3 ………(1)
\nComparing,
\nCoeff. of cos x, 9 = 4\u03bb + 5\u00b5
\n\u21d2 4\u03bb + 5\u00b5 – 9 = 0 ……….(2)
\nCoeff. of sin x, -1 = -5\u03bb + 4\u00b5
\n\u21d2 -5\u03bb + 4\u00b5 + 1 = 0 ………(3)
\nCoeff. of constant, 0 = \u03b3
\nSolving (2) & (3),
\n<\/p>\n
\nEvaluate \\(\\int \\frac{2 \\cos x+3 \\sin x}{4 \\cos x+5 \\sin x} d x\\). [(TS) May ’19, ’16; (AP) Mar. ’18, ’15]
\nSolution:
\nLet 2 cos x + 3 sin x = \u03bb \\(\\frac{d}{dx}\\) (4 cos x + 5 sin x) + \u00b5(4 cos x + 5 sin x) + \u03b3
\n= \u03bb(-4 sin x + 5 cos x) + \u00b5(4 cos x + 5 sin x) + \u03b3 ………(1)
\n= -4\u03bb sin x + 5\u03bb cos x + 4\u00b5 cos x + 5\u00b5 sin x + \u03b3
\nComparing the coefficients of cos x on both sides, we get
\n5\u03bb + 4\u00b5 = 2 ……..(2)
\nComparing the coefficients of sinx on both sides, we get
\n-4\u03bb + 5\u00b5 = 3 ……….(3)
\nSolving (2) and (3)
\n
\nComparing constant terms on both sides, we get X = 0
\nFrom (1),
\n2 cos x + 3 sin x = \\(\\frac{-2}{41}\\) (-4 sin x + 5 cos x) + \\(\\frac{23}{41}\\) (4 cos x + 5 sin x) + 0
\n= \\(\\frac{-2}{41}\\) (-4 sin x + 5 cos x) + \\(\\frac{23}{41}\\) (4 cos x + 5 sin x)
\nNow,
\n
\nPut 4 cos x + 5 sin x = t
\n(-4 sin x + 5 cos x) dx = dt
\n<\/p>\n
\nEvaluate \\(\\int \\frac{2 x+5}{\\sqrt{x^2-2 x+10}} d x\\). [(AP) May ’17; Mar. ’15 (TS)]
\nSolution:
\nWrite 2x + 5 = A . \\(\\frac{d}{dx}\\) (x2 – 2x + 10) + B
\n= A(2x – 2) + B ………(1)
\nCoeff. of x, 2 = 2A then A = 1
\nConstant, 5 = -2A + B then B = 5 + 2 = 7
\nFrom (1), 2x + 5 = (2x – 2) + 7
\n<\/p>\n
\nEvaluate \\(\\int \\frac{x+1}{x^2+3 x+12} d x\\). [(AP) Mar. ’17; May ’16]
\nSolution:
\nWrite x + 1 = A \\(\\frac{d}{dx}\\) [x2 + 3x + 12) + B
\n= A(2x + 3) + B ……..(1)
\nCoefficient of x, 1 = A(2) then A = \\(\\frac{1}{2}\\)
\nConstant, 1 = 3A + B then B = 1 – \\(\\frac{3}{2}\\) = \\(\\frac{-1}{2}\\)
\nFrom (1), x + 1 = \\(\\frac{1}{2}\\)(2x + 3) – \\(\\frac{1}{2}\\)
\n<\/p>\n
\nEvaluate \\(\\int \\sqrt{\\frac{5-x}{x-2}} d x\\). [(AP) May ’19, (TS) ’17]
\nSolution:
\n
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int(6 x+5) \\sqrt{6-2 x^2+x} d x\\). [(AP) & (TS) May ’18]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int x \\sqrt{1+x-x^2} d x\\). [May ’12]
\nSolution:
\nTake x = A \\(\\frac{d}{dx}\\) (1 + x – x2<\/sup>) + B
\nx = A(1 – 2x) + B …….(1)
\n= A – 2Ax + B
\nComparing the coefficient of x on both sides, we get
\n-2A = 1
\n\u21d2 A = \\(-\\frac{1}{2}\\)
\nComparing constant terms on both sides, we get
\nA + B = 0
\n\u21d2 \\(-\\frac{1}{2}\\) + B = 0
\n\u21d2 B = \\(\\frac{1}{2}\\)
\nFrom (1), x = \\(-\\frac{1}{2}\\)(1 – 2x) + \\(\\frac{1}{2}\\)
\n<\/p>\n
\nEvaluate \\(\\int(3 x-2) \\sqrt{2 x^2-x+1} d x\\). [(TS) May ’15; May ’03]
\nSolution:
\nTake 3x – 2 = A \\(\\frac{d}{dx}\\)(2x2<\/sup> – x + 1) + B
\n3x – 2 = A(4x – 1) + B …….(1)
\n3x – 2 = 4Ax – A + B
\nComparing the coefficients of x on both sides, we get
\n4A = 3
\n\u21d2 A = \\(\\frac{3}{4}\\)
\nComparing the constant terms on both sides
\n-A + B = -2
\n\u21d2 \\(-\\frac{3}{4}\\) + B = -2
\n\u21d2 B = \\(\\frac{-5}{4}\\)
\nFrom (1), 3x – 2 = \\(\\frac{3}{4}\\) (4x – 1) – \\(\\frac{5}{4}\\)
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{(1+x) \\sqrt{3+2 x-x^2}}\\). [(TS) Mar. ’20; May ’14, ’05]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{(x+1) \\sqrt{2 x^2+3 x+1}}\\). [Mar. ’18 (TS)]
\nSolution:
\n<\/p>\n
\nEvaluate the Reduction formula for In<\/sub> = \u222bsinn<\/sup>x dx and hence find \u222bsin4<\/sup>x dx, \u222bsin5<\/sup>x dx. [(TS) Mar. ’20, May 18; (AP) May ’19, 15; Mar. ’17; Mar. ’14, ’13]
\nSolution:
\n
\n<\/p>\n
\nObtain the Reduction formula for \u222bcosn<\/sup>x dx for n \u2265 2 and deduce the value of \u222bcos5<\/sup>x dx. [(AP) Mar. ’20, May ’18; (TS) May ’19, ’16; Mar. ’17]
\nSolution:
\n
\n<\/p>\n
\nFind the Reduction formula of \u222btann<\/sup>x dx for an integer n \u2265 2. And deduce the value of \u222btan6<\/sup>x dx. [(AP) Mar. ’18, ’15; May ’16; (TS) ’17; Mar. ’12; May ’13]
\nSolution:
\n
\n<\/p>\n
\nFind the Reduction formula of \u222bcotn<\/sup>x dx for an integer n \u2265 2. And deduce the value of \u222bcot4<\/sup>x dx. [Mar. ’19 (TS); Mar. ’16 (AP); May ’11]
\nSolution:
\n<\/p>\n
\nFind the Reduction formula for \u222bcosecn<\/sup>x dx for an integer n \u2265 2 and deduce the value of \u222bcosec5<\/sup>x dx. [Mar. ’19 (AP); Mar. ’16 (TS); May ’14]
\nSolution:
\n
\n<\/p>\n
\nFind the Reduction formula for \u222bsecn<\/sup>x dx for an integer n \u2265 2 and deduce the value of \u222bsec5<\/sup>x dx. [(AP) May ’17; (TS) May ’15; Mar. ’04]
\nSolution:
\n\\(I_n=\\int \\sec ^n x d x=\\int \\sec ^{n-2} x \\cdot \\sec ^2 x d x\\)
\n<\/p>\n
\nEvaluate \\(\\int \\frac{\\sin 2 x}{a \\cos ^2 x+b \\sin ^2 x} d x\\)
\nSolution:
\nPut a cos2<\/sup>x + b sin2<\/sup>x = t
\nthen [a . 2 cos x (-sin x) + b . 2 sin x . cos x] dx = dt
\n\u21d2 [-a sin 2x + b sin 2x] dx = dt
\n\u21d2 (b – a) sin 2x dx = dt
\n\u21d2 sin 2x dx = \\(\\frac{1}{b-a}\\) dt
\n<\/p>\n
\nEvaluate \u222bx cos-1<\/sup>x dx. [Mar. ’09]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \u222bx sin-1<\/sup>x dx. [Mar. ’04]
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{x^2+x+1}\\)
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{\\sqrt{1+x-x^2}}\\)
\nSolution:
\n<\/p>\n
\nEvaluate \\(\\int \\frac{\\cos x}{\\sin ^2 x+4 \\sin x+5} d x\\). [Mar. ’07, ’03]
\nSolution:
\nPut sin x = t then cos dx = dt
\n<\/p>\n
\nEvaluate \\(\\int \\frac{\\sin x \\cos x}{\\cos ^2 x+3 \\cos x+2} d x\\)
\nSolution:
\nPut cos x = t
\n\u21d2 -sin x dx = dt
\n\u21d2 sin x dx = -dt
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{\\left(x^2+a^2\\right)\\left(x^2+b^2\\right)}\\)
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{\\left(x^2+a^2\\right)^2}\\)
\nSolution:
\n<\/p>\n
\nEvaluate \u222beax<\/sup>\u00a0sin(bx + c) dx. [Mar. ’19 (TS)]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\sqrt{\\mathbf{a}^2-x^2} d x\\). [Mar. ’02]
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\sqrt{1+3 x-x^2} d x\\). [Mar. ’11]
\nSolution:
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{\\sin x+\\sqrt{3} \\cos x}\\)
\nSolution:
\n
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{\\tan ^{-1} x}{x^2} d x\\). [May ’01]
\nSolution:
\n\\(\\int \\frac{\\tan ^{-1} x}{x^2} d x=\\int \\tan ^{-1} x \\cdot x^{-2} d x\\)
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{a \\sin x+b \\cos x} d x\\). [May ’03]
\nSolution:
\nLet a = r cos \u03b8, b = r sin \u03b8 then r = \\(\\sqrt{a^2+b^2}\\)
\nNow a sin x + b cos x = r cos \u03b8 sin x + r sin \u03b8 cos x = r[sin (x + \u03b8)]
\n<\/p>\n
\nEvaluate \u222bex<\/sup> log(e2x<\/sup> + 5ex<\/sup> + 6) dx.
\nSolution:
\n
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}} d x\\)
\nSolution:
\nPut x = t2<\/sup>, then dx = 2t dt
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{(x-a)(x-b)(x-c)} d x\\)
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{x(x+1)(x+2)}\\)
\nSolution:
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{7 x-4}{(x-1)^2(x+2)} d x\\)
\nSolution:
\nLet \\(\\frac{7 x-4}{(x-1)^2(x+2)}=\\frac{A}{x-1}+\\frac{B}{(x-1)^2}+\\frac{C}{x+2}\\)
\n\u21d2 \\(\\frac{7 x-4}{(x-1)^2(x+2)}=\\frac{A(x-1)(x+2)+B(x+2)+C(x-1)^2}{(x-1)^2(x+2)}\\)
\n\u21d2 7x – 4 = A(x – 1) (x + 2) + B(x + 2) + C(x – 1)2<\/sup>
\nIf x = 1,
\n1 = 7(1) – 4 = B(1 + 2)
\n\u21d2 3 = 3B
\n\u21d2 B = 1
\nIf x = -2,
\n7(-2) – 4 = C(-2 – 1)2<\/sup>
\n\u21d2 -14 – 4 = C(-3)2<\/sup>
\n\u21d2 -18 = 9C
\n\u21d2 C = -2
\n(1) \u21d2 7x – 4 = Ax2<\/sup> + Ax – 2A + Bx + 2B + Cx2<\/sup> – 2Cx + C
\nComparing the coefficients of x2<\/sup> on both sides, we get
\nA + C = 0
\n\u21d2 A – 2 = 0
\n\u21d2 A = 2
\n<\/p>\n
\nEvaluate \\(\\int \\frac{x^2}{(x+1)(x+2)^2} d x\\)
\nSolution:
\nLet \\(\\frac{x^2}{(x+1)(x+2)^2}=\\frac{A}{x+1}+\\frac{B}{x+2}+\\frac{C}{(x+2)^2}\\)
\n\u21d2 \\(\\frac{x^2}{(x+1)(x+2)^2}=\\frac{A(x+2)^2+B(x+1)(x+2)+C(x+1)}{(x+1)(x+2)^2}\\)
\n\u21d2 x2<\/sup> = A(x + 2)2<\/sup> + B(x + 1)(x + 2) + C(x + 1) …..(1)
\nIf x = -1, then (-1)2<\/sup> = A(-1 + 2)2<\/sup>
\n\u21d2 1 = A(1)2<\/sup>
\n\u21d2 A = 1
\nIf x = -2, then (-2) = C(-2 + 1)
\n\u21d2 4 = C(-1)
\n\u21d2 4 = -C
\n\u21d2 C = -4
\n(1) \u21d2 x2<\/sup> = Ax2<\/sup> + 4Ax + 4A + Bx2<\/sup> + 3Bx + 2B + Cx + C
\nComparing the coefficients of x2<\/sup> on both sides, we get
\n1 = A + B
\n\u21d2 1 + B = 1
\n\u21d2 B = 0
\n<\/p>\n
\nEvaluate \\(\\int \\frac{2 x+3}{x^3+x^2-2 x} d x\\)
\nSolution:
\nGiven \\(\\int \\frac{2 x+3}{x^3+x^2-2 x} d x=\\int \\frac{2 x+3}{x\\left(x^2+x-2\\right)} d x=\\int \\frac{2 x+3}{x(x+2)(x-1)} d x\\)
\nLet \\(\\frac{2 x+3}{x(x+2)(x-1)}=\\frac{A}{x}+\\frac{B}{x+2}+\\frac{C}{x-1}\\)
\n\u21d2 \\(\\frac{2 x+3}{x(x+2)(x-1)}=\\frac{A(x+2)(x-1)+B x(x-1)+C x(x+2)}{x(x+2)(x-1)}\\)
\n\u21d2 2x + 3 = A(x + 2)(x – 1) + B(x – 1) x + Cx(x + 2)
\nIf x = 0, then 2(0) + 3 = A(0 + 2) (0 – 1)
\n\u21d2 3 = A(2)(-1)
\n\u21d2 3 = -2A
\n\u21d2 A = \\(\\frac{-3}{2}\\)
\nIf x = -2, then 2(-2) + 3 = B(-2 – 1) (-2)
\n\u21d2 -4 + 3 = B(-2)(-3)
\n\u21d2 -1 = 6B
\n\u21d2 B = \\(\\frac{-1}{6}\\)
\nIf x = 1, then 2(1) + 3 = C(1)(1 + 2)
\n\u21d2 5 = C(1)(3)
\n\u21d2 3C = 5
\n\u21d2 C = \\(\\frac{5}{3}\\)
\n<\/p>\n
\nEvaluate \\(\\int \\frac{x+3}{(x-1)\\left(x^2+1\\right)} d x\\). [May ’07]
\nSolution:
\nLet \\(\\frac{x+3}{(x-1)\\left(x^2+1\\right)}=\\frac{A}{x-1}+\\frac{B x+C}{x^2+1}\\)
\n\\(\\frac{x+3}{(x-1)\\left(x^2+1\\right)}=\\frac{A\\left(x^2+1\\right)+(B x+C)(x-1)}{(x-1)\\left(x^2+1\\right)}\\)
\nx + 3 = A(x2<\/sup> + 1) + (Bx + C) (x – 1) ………(1)
\nIf x = 1 then 1 + 3 = A(12<\/sup> + 1)
\n\u21d2 4 = A(2)
\n\u21d2 A = 2
\nfrom (1), x + 3 = Ax2<\/sup> + A + Bx2<\/sup> – Bx + Cx – C
\nComparing x2<\/sup> coefficients on both sides, we get
\nA + B = 0
\n\u21d2 2 + B = 0
\n\u21d2 B = -2
\nComparing coefficients of x on both sides, we get
\n-B + C = 1
\n\u21d2 -(-2) + C = 1
\n\u21d2 2 + C = 1
\n\u21d2 C = 1 – 2
\n\u21d2 C = -1
\n<\/p>\n
\nEvaluate \\(\\int \\frac{2 x+3}{(x+3)\\left(x^2+4\\right)} d x\\). [May ’02]
\nSolution:
\nLet \\(\\frac{2 x+3}{(x+3)\\left(x^3+4\\right)}=\\frac{A}{x+3}+\\frac{B x+C}{x^2+4}\\)
\n\\(\\frac{2 x+3}{(x+2)\\left(x^2+4\\right)}=\\frac{A\\left(x^2+4\\right)+(B x+C)(x+3)}{(x+3)\\left(x^2+4\\right)}\\)
\n2x + 3 = A(x2<\/sup> + 4) + (Bx + C)(x + 3) …….(1)
\nIf x = -3 then
\n2(-3) + 3 = A[(-3)2<\/sup> + 4]
\n\u21d2 -6 + 3 = A(9 + 4)
\n\u21d2 13A = -3
\n\u21d2 A = \\(\\frac{-3}{13}\\)
\nFrom (1),
\n2x + 3 = Ax2<\/sup> + 4A + Bx2<\/sup> + 3Bx + Cx + 3C
\nComparing x2<\/sup> coefficients on both sides, we get
\nA + B = 0
\n\u21d2 \\(\\frac{-3}{13}\\) + B = 0
\n\u21d2 B = \\(\\frac{3}{13}\\)
\nComparing x coefficients on both sides, we get
\n3B + C = 2
\n\u21d2 3(\\(\\frac{3}{13}\\)) + C = 2
\n\u21d2 C = 2 – \\(\\frac{3}{13}\\)
\n\u21d2 C = \\(\\frac{17}{13}\\)
\n
\n<\/p>\n
\nEvaluate \\(\\int \\frac{1}{(1-x)\\left(4+x^2\\right)} d x\\)
\nSolution:
\n\\(\\frac{1}{(1-x)\\left(4+x^2\\right)}=\\frac{A}{1-x}+\\frac{B x+C}{x^2+4}\\)
\n1 = A(x2<\/sup> + 4) + (Bx + C)(1 – x)
\nPut x = 1 then A = \\(\\frac{1}{5}\\)
\nCoeff. of x2<\/sup>, 0 = A – B \u21d2 B = \\(\\frac{1}{5}\\)
\nConstant, 1 = C + 4A \u21d2 C = \\(\\frac{1}{5}\\)
\n<\/p>\n
\nEvaluate \\(\\int \\frac{d x}{x^3+1}\\). [May ’03]
\nSolution:
\n\\(\\frac{1}{x^3+1}=\\frac{1}{(x+1)\\left(x^2-x+1\\right)}=\\frac{A}{x+1}+\\frac{B x+C}{x^2-x+1}\\) ……(1)
\n1 = A(x2<\/sup> – x + 1) + (Bx + C) (x + 1)
\nPut x = -1, 1 = A(3) \u21d2 A = \\(\\frac{1}{3}\\)
\nCoeff. of x2<\/sup>, 0 = A + B \u21d2 B = \\(\\frac{-1}{3}\\)
\nConstant, 1 = A + C \u21d2 C = 1 – \\(\\frac{1}{3}\\) = \\(\\frac{2}{3}\\)
\n
\n<\/p>\n
\nEvaluate \\(\\int \\tan ^{-1} \\sqrt{\\frac{1-x}{1+x}} d x\\)
\nSolution:
\n
\n<\/p>\n
\nFind the Reduction formula for \u222bsinm<\/sup>x cosn<\/sup>x dx for a +ve integer and n \u2265 2.
\nSolution:
\n
\n<\/p>\n
\nIf In<\/sub> = \u222b(log x)n<\/sup> dx, then show that In<\/sub> = x(log x)n<\/sup> – nIn<\/sub> – 1 and find \u222b(log x)4<\/sup> dx.
\nSolution:
\n
\nNow I4<\/sub> = \u222b(log x)4<\/sup> dx
\n= x(log x)4<\/sup> – 4I3<\/sub>
\n= x(log x)4<\/sup> – 4[x(log x)3<\/sup> – 4I2<\/sub>]
\n= x(log x)4<\/sup> – 4x(log x)3<\/sup> + 16I2<\/sub>
\n= x(log x)4<\/sup> – 4x(log x)3<\/sup> + 16[x(log x)2<\/sup> – 2I1<\/sub>]
\n= x(log x)4<\/sup> – 4x(log x)3<\/sup> + 16x(log x)2<\/sup> – 32I1<\/sub>
\n= x(log x)4<\/sup> – 4x(log x)3<\/sup> + 16x(log x)2<\/sup> – 32[x(log x) – x] + c
\n= x(log x)4<\/sup> – 4x(log x)3<\/sup> + 16x(log x)2<\/sup> – 32x log x + 32x + c<\/p>\n","protected":false},"excerpt":{"rendered":"