{"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

TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type<\/h2>\n

Question 1.
\nEvaluate \\(\\int \\frac{d x}{4+5 \\sin x}\\). [Mar. ’05]
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 2.
\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\"TS
\n\"TS<\/p>\n

Question 3.
\nEvaluate \\(\\int \\frac{1}{1+\\sin x+\\cos x} d x\\). [(AP) Mar. ’20; (TS) ’15]
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 4.
\nEvaluate \\(\\int \\frac{d x}{4 \\cos x+3 \\sin x}\\). [(TS) Mar. ’18]
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 5.
\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\"TS
\n\"TS<\/p>\n

Question 6.
\nEvaluate \\(\\int \\frac{1}{2-3 \\cos 2 x} d x\\). [May ’10, ’05]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 7.
\nEvaluate \\(\\int \\frac{d x}{5+4 \\cos 2 x}\\). [May ’13, Mar. ’11]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 8.
\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\"TS
\nComparing constant terms on both sides, we get
\n\u00b5 + \u03b3 = 7
\n\u21d2 2 + \u03b3 = 7
\n\u21d2 \u03b3 = 5
\n\"TS
\n\"TS
\n\"TS
\nwhere k is an integration constant.<\/p>\n

\"TS<\/p>\n

Question 9.
\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\"TS
\n\"TS
\n\"TS<\/p>\n

Question 10.
\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\"TS<\/p>\n

Question 11.
\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\"TS
\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\"TS
\nPut 4 cos x + 5 sin x = t
\n(-4 sin x + 5 cos x) dx = dt
\n\"TS<\/p>\n

Question 12.
\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\"TS<\/p>\n

\"TS<\/p>\n

Question 13.
\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\"TS<\/p>\n

Question 14.
\nEvaluate \\(\\int \\sqrt{\\frac{5-x}{x-2}} d x\\). [(AP) May ’19, (TS) ’17]
\nSolution:
\n\"TS
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 15.
\nEvaluate \\(\\int(6 x+5) \\sqrt{6-2 x^2+x} d x\\). [(AP) & (TS) May ’18]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 16.
\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\"TS<\/p>\n

Question 17.
\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\"TS<\/p>\n

Question 18.
\nEvaluate \\(\\int \\frac{d x}{(1+x) \\sqrt{3+2 x-x^2}}\\). [(TS) Mar. ’20; May ’14, ’05]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 19.
\nEvaluate \\(\\int \\frac{d x}{(x+1) \\sqrt{2 x^2+3 x+1}}\\). [Mar. ’18 (TS)]
\nSolution:
\n\"TS<\/p>\n

Question 20.
\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\"TS
\n\"TS<\/p>\n

Question 21.
\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\"TS
\n\"TS<\/p>\n

Question 22.
\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\"TS
\n\"TS<\/p>\n

Question 23.
\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\"TS<\/p>\n

Question 24.
\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\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 25.
\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\"TS<\/p>\n

Question 26.
\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\"TS<\/p>\n

Question 27.
\nEvaluate \u222bx cos-1<\/sup>x dx. [Mar. ’09]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 28.
\nEvaluate \u222bx sin-1<\/sup>x dx. [Mar. ’04]
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 29.
\nEvaluate \\(\\int \\frac{d x}{x^2+x+1}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 30.
\nEvaluate \\(\\int \\frac{d x}{\\sqrt{1+x-x^2}}\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 31.
\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\"TS<\/p>\n

Question 32.
\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\"TS
\n\"TS<\/p>\n

Question 33.
\nEvaluate \\(\\int \\frac{d x}{\\left(x^2+a^2\\right)\\left(x^2+b^2\\right)}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 34.
\nEvaluate \\(\\int \\frac{d x}{\\left(x^2+a^2\\right)^2}\\)
\nSolution:
\n\"TS<\/p>\n

Question 35.
\nEvaluate \u222beax<\/sup>\u00a0sin(bx + c) dx. [Mar. ’19 (TS)]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 36.
\nEvaluate \\(\\int \\sqrt{\\mathbf{a}^2-x^2} d x\\). [Mar. ’02]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 37.
\nEvaluate \\(\\int \\sqrt{1+3 x-x^2} d x\\). [Mar. ’11]
\nSolution:
\n\"TS<\/p>\n

Question 38.
\nEvaluate \\(\\int \\frac{d x}{\\sin x+\\sqrt{3} \\cos x}\\)
\nSolution:
\n\"TS
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 39.
\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\"TS<\/p>\n

Question 40.
\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\"TS<\/p>\n

Question 41.
\nEvaluate \u222bex<\/sup> log(e2x<\/sup> + 5ex<\/sup> + 6) dx.
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 42.
\nEvaluate \\(\\int \\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}} d x\\)
\nSolution:
\nPut x = t2<\/sup>, then dx = 2t dt
\n\"TS
\n\"TS<\/p>\n

Question 43.
\nEvaluate \\(\\int \\frac{1}{(x-a)(x-b)(x-c)} d x\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 44.
\nEvaluate \\(\\int \\frac{d x}{x(x+1)(x+2)}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 45.
\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\"TS<\/p>\n

\"TS<\/p>\n

Question 46.
\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\"TS<\/p>\n

Question 47.
\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\"TS<\/p>\n

Question 48.
\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\"TS<\/p>\n

\"TS<\/p>\n

Question 49.
\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\"TS
\n\"TS<\/p>\n

Question 50.
\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\"TS<\/p>\n

Question 51.
\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\"TS
\n\"TS<\/p>\n

Question 52.
\nEvaluate \\(\\int \\tan ^{-1} \\sqrt{\\frac{1-x}{1+x}} d x\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 53.
\nFind the Reduction formula for \u222bsinm<\/sup>x cosn<\/sup>x dx for a +ve integer and n \u2265 2.
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 54.
\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\"TS
\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":"

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 - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/maths-2b-integration-important-questions-long-answer-type\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter Second Year Maths 2B Integration Important Questions Long Answer Type - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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