{"id":22137,"date":"2024-01-11T16:44:32","date_gmt":"2024-01-11T11:14:32","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22137"},"modified":"2024-01-12T16:18:25","modified_gmt":"2024-01-12T10:48:25","slug":"inter-2nd-year-maths-2b-integration-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2b-integration-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2B Integration Important Questions"},"content":{"rendered":"

Students get through Maths 2B Important Questions<\/a> Inter 2nd Year Maths 2B Integration Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2B Integration Important Questions<\/h2>\n

Question 1.
\n\\(\\int\\left(\\frac{1}{1-x^{2}}+\\frac{2}{1+x^{2}}\\right)\\) [May 11]
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\n\\(\\int\\) sec2<\/sup>x cosec2<\/sup>x dx on I \u2282 R \\ ({n\u03c0 : n \u2208 Z} \u222a {(2n + 1) \\(\\frac{\\pi}{2}\\) : n \u2208 Z}) [T.S. Mar. 16; Mar, May 07]
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\n\\(\\int \\frac{1+\\cos ^{2} x}{1-\\cos 2 x}\\) dx on I \u2282 R \\ {n\u03c0 : n \u2208 Z} [Mar. 13]
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\n\\(\\int \\sqrt{1-\\cos 2 x}\\) dx on I \u2282 [2n\u03c0, (n + 1) \u03c0], n \u2208 Z [May 06]
\nSolution:
\n\\(\\int \\sqrt{1-\\cos 2 x}\\) dx = \\(\\int \\sqrt{2}\\) sin x dx
\n= –\\(\\sqrt{2}\\) cos x + C<\/p>\n

Question 5.
\n\\(\\int \\frac{1}{\\cosh x+\\sinh x}\\) dx on R. [A.P. Mar. 16]
\nSolution:
\n\\(\\int \\frac{1}{\\cosh x+\\sinh x}\\) dx
\n= \\(\\int \\frac{\\cosh x-\\sinh x}{\\cosh ^{2} x-\\sinh ^{2} x}\\) dx
\n= \\(\\int\\) (cosh x – sinh x) dx
\n= sinh x – cosh x + C<\/p>\n

\"Inter<\/p>\n

Question 6.
\n\\(\\int \\frac{1}{1+\\cos x}\\) dx on I \u2282 R \\ {(2n + 1)\u03c0 : n \u2208 Z} [T.S. Mar. 15]
\nSolution:
\n\"Inter<\/p>\n

Question 7.
\n\\(\\int \\frac{\\sin \\left(\\tan ^{-1} x\\right)}{1+x^{2}}\\)dx, x \u2208 R. [A.P. Mar. 15]
\nSolution:
\n\\(\\int \\frac{\\sin \\left(\\tan ^{-1} x\\right)}{1+x^{2}}\\)dx
\nt = tan-1<\/sup> x \u21d2 dt = \\(\\frac{d x}{1+x^{2}}\\)
\n\\(\\int \\frac{\\sin \\left(\\tan ^{-1} x\\right)}{1+x^{2}}\\)dx = \\(\\int\\) sin t dt
\n= – cos t + t
\n= -cos (tan-1<\/sup> x) + C<\/p>\n

Question 8.
\n\\(\\int \\frac{\\log (1+x)}{1+x}\\)dx on (-1, \u221e) [T.S. Mar. 15]
\nSolution:
\n\\(\\int \\frac{\\log (1+x)}{1+x}\\)dx
\nt = 1 + x \u21d2 dt = dx
\n\"Inter<\/p>\n

Question 9.
\n\\(\\int \\frac{x^{2}}{\\sqrt{1-x^{6}}}\\) dx on I = (-1, 1). [May 05]
\nSolution:
\n\\(\\int \\frac{x^{2}}{\\sqrt{1-x^{6}}}\\)
\nt = x3<\/sup> \u21d2 dt = mx2<\/sup> dx
\n\\(\\int \\frac{x^{2}}{\\sqrt{1-x^{6}}}\\) = \\(\\frac{1}{3} \\int \\frac{d \\mathrm{t}}{\\sqrt{1-\\mathrm{t}^{2}}}\\)
\n= \\(\\frac{1}{3}\\) sin-1<\/sup> t + C
\n= \\(\\frac{1}{3}\\) sin-1<\/sup> (x3<\/sup>) + C<\/p>\n

\"Inter<\/p>\n

Question 10.
\n\\(\\int \\frac{x^{8}}{1+x^{18}}\\) dx on R. [A.P. Mar. 16]
\nSolution:
\nt = x9<\/sup> \u21d2 dt = 9x8<\/sup> dx
\n\\(\\int \\frac{x^{8} d x}{1+x^{18}}=\\int \\frac{x^{8}}{1+\\left(x^{9}\\right)^{2}} d x\\)
\n= \\(\\frac{1}{9} \\int \\frac{d t}{1+t^{2}}\\) = = \\(\\frac{1}{9}\\) tan-1<\/sup> t + C
\n= \\(\\frac{1}{9}\\) tan-1<\/sup> (x9<\/sup>) + C<\/p>\n

Question 11.
\n\\(\\int \\frac{1}{x \\log x[\\log (\\log x)]}\\) dx on (1, \u221e) [Mar. 11]
\nSolution:
\nt = log (log x)
\ndt = \\(\\frac{1}{\\log x} \\cdot \\frac{1}{x}\\) dx
\n\\(\\int \\frac{1}{x \\log x[\\log (\\log x)]}\\) dx = \\(\\int \\frac{d t}{t}\\)
\n= log |t| + C
\n= log |log(log x)| + C<\/p>\n

Question 12.
\n\\(\\int \\frac{1}{(x+3) \\sqrt{x+2}}\\)dx on I \u2282 (-2, \u221e) [Mar. 14]
\nSolution:
\nx + 2 = t2<\/sup>
\ndx = 2t dt
\n\\(\\int \\frac{d x}{(x+3) \\sqrt{x+2}}=\\int \\frac{2 t d t}{t\\left(t^{2}+1\\right)}\\)
\n= \\(2 \\int \\frac{d t}{t^{2}+1}\\)
\n= 2 tan-1<\/sup> (t) + C
\n= 2 tan-1<\/sup> (\\(\\sqrt{x+2}\\)) + C<\/p>\n

\"Inter<\/p>\n

Question 13.
\n\\(\\int \\frac{\\cot (\\log x)}{x}\\) dx, x \u2208 I \u2282 (0, \u221e) \\ {en\u03c0<\/sup> : n \u2208 Z). [Mar. 05]
\nSolution:
\nt = log x \u21d2 dt = \\(\\frac{\\mathrm{dx}}{\\mathrm{x}}\\) \\(\\int\\)
\n\\(\\int \\frac{\\cot (\\log x)}{x}\\) dx = \\(\\int\\) cot t dt = log (sin t) + C
\n= log |sin (log x)| + C<\/p>\n

Question 14.
\n\\(\\int\\) (tan x + log sec x)ex<\/sup> dx on ((2n – \\(\\frac{1}{2}\\))\u03c0, (2n + \\(\\frac{1}{2}\\))\u03c0) n \u2208 Z. [May 07, Mar. 08]
\nSolution:
\nt = log |sec x| \u21d2 dt = \\(\\frac{1}{\\sec x}\\) . sec x . tan x dx
\n= tan x dx
\n\\(\\int\\) (tan x + log sec x)ex<\/sup> dx = ex<\/sup> . log|sec x| + C<\/p>\n

Question 15.
\n\\(\\int \\sqrt{x}\\) log x dx on (0, \u221e) [T.S. Mar. 16]
\nSolution:
\n\\(\\int \\sqrt{x}\\) log x dx
\n= (log x) \\(\\frac{2}{3}\\) x3\/2<\/sup> – \\(\\frac{2}{3}\\) \\(\\int\\) x3\/2<\/sup> \\(\\frac{1}{x}\\) dx
\n= \\(\\frac{2}{3}\\) x3\/2<\/sup> (log x) – \\(\\frac{2}{3}\\) \\(\\int\\) x1\/2<\/sup> dx
\n= \\(\\frac{2}{3}\\) x3\/2<\/sup> (log x) – \\(\\frac{2}{3}\\) \\(\\frac{x^{3 \/ 2}}{3 \/ 2}\\) + C
\n= \\(\\frac{2}{3}\\) x3\/2<\/sup> (log x) – \\(\\frac{4}{9}\\) x3\/2<\/sup> + C<\/p>\n

Question 16.
\n\\(\\int\\) ex<\/sup> (tan x + sec2<\/sup> x)dx on I \u2282 R \\ {(2n + 1)\\(\\frac{\\pi}{2}\\) : n \u2208 Z} [Mar 06]
\nSolution:
\nf(x) = tanx = f'(x) \u21d2 sec2<\/sup> x dx
\nI = \\(\\int\\) ex<\/sup> [f(x) + f'(x)] dx = ex<\/sup>. f(x) + C
\n= ex<\/sup> . tan x + C<\/p>\n

\"Inter<\/p>\n

Question 17.
\n\\(\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right)\\) dx on (0, \u221e) [A.P. Mar. 15, Mar. 13]
\nSolution:
\n\\(\\int e^{x}\\left(\\frac{1+x \\log x}{x}\\right)\\) dx = \\(\\int\\) ex<\/sup> (log x + \\(\\frac{1}{x}\\))dx
\n= ex<\/sup> . log x + C<\/p>\n

Question 18.
\n\\(\\int \\frac{\\cos x}{\\sin ^{2} x+4 \\sin x+5}\\)dx [Mar. 07]
\nSolution:
\nt = sin x \u21d2 dt = cos x dx
\nI = \\(\\int \\frac{d t}{t^{2}+4 t+5}=\\int \\frac{d t}{(t+2)^{2}+1}\\)
\n= tan-1<\/sup>(t + 2) + C
\n= tan-1<\/sup>(sin x + 2) + C<\/p>\n

Question 19.
\n\\(\\int \\frac{d x}{(x+1)(x+2)}\\) [Mar. 14, May 11]
\nSolution:
\n\"Inter<\/p>\n

Question 20.
\n\\(\\int \\frac{e^{x}(1+x)}{\\cos ^{2}\\left(x e^{x}\\right)}\\) dx on I \u2282 R \\ {x \u2208 R : cos (xex<\/sup>) = 0} [T.S. Mar. 17]
\nSolution:
\nt = x . ex<\/sup>
\ndt = (x . ex<\/sup> + ex<\/sup>) dx = ex<\/sup> (1 + x) dx
\n\\(\\int \\frac{e^{x}(1+x)}{\\cos ^{2}\\left(x e^{x}\\right)}\\) dx = \\(\\int \\frac{d t}{\\cos ^{2} t}=\\int \\sec ^{2} t d t\\)
\n= tan t + C
\n= tan (x . ex<\/sup>) + C<\/p>\n

Question 21.
\n\\(\\int\\) x tan-1<\/sup> x dx, x\u2208 R [Mar. 05]
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 22.
\n\\(\\int \\sqrt{1+3 x-x^{2}} d x\\) [May 11]
\nSolution:
\n\\(\\int \\sqrt{1+3 x-x^{2}} d x=\\int \\sqrt{1-\\left(x^{2}-3 x\\right)} d x\\)
\n= \\(\\int \\sqrt{1-\\left(x^{2}-3 x+\\frac{9}{4}-\\frac{9}{4}\\right)} d x\\)
\n\"Inter<\/p>\n

Question 23.
\n\\(\\int \\frac{9 \\cos x-\\sin x}{4 \\sin x+5 \\cos x} d x\\) [T.S. Mar. 17; Mar. 08]
\nSolution:
\n\"Inter<\/p>\n

Question 24.
\n\\(\\int \\frac{d x}{5+4 \\cos 2 x}\\) [Mar. 11]
\nSolution:
\n\"Inter<\/p>\n

Question 25.
\nObtain reduction formula for In<\/sub> = \\(\\int\\) cotn<\/sup> x dx, n being a positive integer. n \u2265 2 and deduce the value of \\(\\int\\) cot4<\/sup> x dx. [T.S. Mar. 19] [A.P. Mar. 16; May 11]
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 26.
\nObtain reduction formula for In<\/sub> = \\(\\int\\) cosecn<\/sup> x dx, n being a positive integer. n \u2265 2 and deduce the value of \\(\\int\\) cosec5<\/sup> x dx. [T.S. Mar. 16]
\nSolution:
\nIn<\/sub> = \\(\\int\\) cosecn<\/sup> x dx = \\(\\int\\) cosecn – 2<\/sup>x . cosec2<\/sup> x dx
\n= cosecn – 2<\/sup>x (-cot x) + \\(\\int\\) cot x . (n – 2) cosecn – 3<\/sup> . (cot x) dx
\n= – cosecn – 2<\/sup>x . cot x + (n – 2) \\(\\int\\) cosecn – 2<\/sup> x . (cosec2<\/sup> x -1) dx
\n= – cosecn – 2<\/sup>x . cot x + (n – 2) In – 2<\/sub> – (n – 2)In<\/sub>
\nIn<\/sub> (1 + n – 2) = – cosecn – 2<\/sup> x . cot x + (n – 2) In – 2<\/sub>
\n\"Inter<\/p>\n

Question 27.
\nEvaluate \\(\\int \\frac{2 x+5}{\\sqrt{x^{2}-2 x+10}}\\) dx. [T.S. Mar. 15]
\nSolution:
\nWe write
\n2x + 5 = A \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (x2<\/sup> – 2x + 10) + B
\n= A(2x – 2) + B
\nOn comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
\nThus 2x + 5 = (2x – 2) + 7
\nHence \\(\\int \\frac{2 x+5}{\\sqrt{x^{2}-2 x+10}}\\) dx
\n= \\(\\int \\frac{2 x-2}{\\sqrt{x^{2}-2 x+10}} d x+7 \\int \\frac{d x}{\\sqrt{x^{2}-2 x+10}}+C\\)
\n= \\(2 \\sqrt{x^{2}-2 x+10}+7 \\int \\frac{d x}{\\sqrt{(x-1)^{2}+3^{2}}}\\) + C
\n= \\(2 \\sqrt{x^{2}-2 x+10}+7 \\sinh ^{-1}\\left(\\frac{x-1}{3}\\right)\\) + C<\/p>\n

Question 28.
\nEvaluate \\(\\int\\) sin4<\/sup> x dx. [Mar. 14]
\nSolution:
\nIn<\/sub> = sinn<\/sup> x dx = \\(-\\frac{\\sin ^{n-1} x \\cdot \\cos x}{n}+\\frac{n-1}{n} \\cdot I_{n-2}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 29.
\n\\(\\int \\frac{2 \\cos x+3 \\sin x}{4 \\cos x+5 \\sin x} d x\\) [A.P. Mar. 15]
\nSolution:
\nLet 2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
\nEquating the coefficient of sin x and cos x we get
\n4A + 5B = 2
\n5A – 4B = 3
\n\"Inter<\/p>\n

Question 30
\n\\(\\int \\frac{1}{1+\\sin x+\\cos x}\\) dx [T.S. Mar. 15]
\nSolution:
\n\"Inter<\/p>\n

Question 31.
\n\\(\\int\\) (6x + 5) \\(\\sqrt{6-2 x^{2}+x}\\) dx [May 06]
\nSolution:
\nLet 6x + 5 = A(1 – 4x) + B
\nEquating the co-efficients of x
\n6 = -4 A \u21d2 A = \\(\\frac{-3}{2}\\)
\nEquating the constants
\nA + B = 5
\nB = 5 – A = 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 32.
\n\\(\\int \\frac{d x}{4+5 \\sin x}\\) [Mar. 05]
\nSolution:
\nt = tan \\(\\frac{x}{2}\\) \u21d2 dt = sec2<\/sup> \\(\\frac{x}{2}\\) . \\(\\frac{1}{2}\\) dx
\n\"Inter
\n\"Inter<\/p>\n

Question 33.
\n\\(\\int \\frac{d x}{(1+x) \\sqrt{3+2 x-x^{2}}}\\) [May 05]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 34.
\n\\(\\int \\frac{d x}{4 \\cos x+3 \\sin x}\\) [Mar. 06]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 35.
\n\\(\\int \\frac{2 \\sin x+3 \\cos x+4}{3 \\sin x+4 \\cos x+5} d x\\) [Mar. 14, 11] [A.P. & T.S. Mar. 16]
\nSolution:
\nLet 2 sin x + 3 cos x + 4
\n= A(3 sin x+4 cos x + 5) + 3(3 cos x – 4sin x) + C
\nEquating the co-efficients of
\nsin x. we get 3A – 4B = 2
\ncos x, we get 4A + 3B = 3
\n\"Inter
\n\"Inter
\nSubstituting in (1)
\nI = \\(\\frac{18}{25}\\) . x + \\(\\frac{1}{25}\\) log |3 sin x + 4 cos x + 5| – \\(\\frac{4}{5\\left(3+\\tan \\frac{x}{2}\\right)}\\) + C<\/p>\n

Question 36.
\n\\(\\int \\frac{x+3}{(x-1)\\left(x^{2}+1\\right)} d x\\) dx [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\u21d2 (x + 3) = A(x2<\/sup> + 1) + (Bx + C)(x – 1) ………………… (1)
\nPut x = 1 in (1)
\nThen 4 = A(1 + 1) + 0 \u21d2 A = 2
\nPut x = 0 in (1)
\n3 = A(1) + C(-1)
\n\u21d2 A – C = 3 \u21d2 C = A – 3 = 2 – 3 = -1
\nEquating coefficient of x2<\/sup> in (1)
\n0 = A + B
\n\u21d2 B = -A = -2
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 37.
\nFind \\(\\int \\frac{d x}{3 \\cos x+4 \\sin x+6}\\) [Mar. 13]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 38.
\nFind \\(\\int\\) 2x7<\/sup> dx on R.
\nSolution:
\n\\(\\int\\) 2x7<\/sup> dx = 2 \\(\\int\\) x7<\/sup> dx
\n= 2 . \\(\\frac{x^{8}}{8}\\) + C
\n= \\(\\frac{x^{8}}{4}\\) + C<\/p>\n

Question 39.
\nEvaluate \\(\\int\\) cot2<\/sup>x dx on I \u2282 R \\ {n\u03c0 : n \u2208 Z}.
\nSolution:
\n\\(\\int\\) cot2<\/sup>x dx = \\(\\int\\) (cosec2<\/sup>x – 1) dx
\n= \\(\\int\\) cosec2<\/sup> x dx – \\(\\int\\) dx
\n= -cot x – x + C<\/p>\n

Question 40.
\nEvaluate \\(\\int\\left(\\frac{x^{6}-1}{1+x^{2}}\\right)\\) dx for x \u2208 R.
\nSolution:
\n\\(\\int\\left(\\frac{x^{6}-1}{1+x^{2}}\\right)\\) dx = \\(\\int\\)[(x4<\/sup> – x2<\/sup> + 1) – \\(\\frac{2}{1+x^{2}}\\)] dx
\n= \\(\\int\\) x4<\/sup> dx – \\(\\int\\) x2<\/sup> dx + \\(\\int\\) dx – 2 \\(\\int \\frac{d x}{1+x^{2}}\\)
\n= \\(\\frac{x^{5}}{5}\\) – \\(\\frac{x^{3}}{3}\\) + x – 2 tan-1<\/sup> x + C.<\/p>\n

\"Inter<\/p>\n

Question 41.
\nFind \\(\\int\\) (1 – x) (4 – 3x) (3 + 2x) dx, x \u2208 R.
\nSolution:
\n(1 – x) (4 – 3x) (3 + 2x) = 6x3<\/sup> – 5x2<\/sup> – 13x + 12
\n\\(\\int\\)(1 – x) (4 – 3x) (3 + 2x) dx
\n= \\(\\int\\) (6x3<\/sup> – 5x2<\/sup> – 13x + 12) dx
\n= 6\\(\\int\\) x3<\/sup>dx – 5 \\(\\int\\) x2<\/sup> dx – 13 \\(\\int\\) x dx + 12 \\(\\int\\) dx
\n= \\(\\frac{6 x^{4}}{4}\\) – 5 \\(\\frac{x^{3}}{3}\\) – \\(\\frac{13 x^{2}}{2}\\) + 12x + C
\n= \\(\\frac{3}{2}\\)x4<\/sup> – \\(\\frac{5}{3}\\)x3<\/sup> – \\(\\frac{13}{2}\\)x2<\/sup> + 12x + C.<\/p>\n

Question 42.
\nEvaluate \\(\\int\\left(x+\\frac{1}{x}\\right)^{3}\\) dx, x > 0.
\nSolution:
\n(x + \\(\\frac{1}{x}\\))3<\/sup> = x3<\/sup> + 3x + \\(\\frac{3}{x}\\) + \\(\\frac{1}{x^{3}}\\)
\n\\(\\int\\) (x + + \\(\\frac{1}{x}\\))3<\/sup> dx = \\(\\int\\) (x3<\/sup> + 3x + \\(\\frac{3}{x}\\) + \\(\\frac{1}{x^{3}}\\)) dx
\n= \\(\\int\\) x3<\/sup> dx + 3 \\(\\int\\) x dx + 3 \\(\\int\\) \\(\\frac{\\mathrm{dx}}{\\mathrm{x}}\\) + \\(\\int\\) \\(\\frac{d x}{x^{3}}\\)
\n= \\(\\frac{x^{4}}{4}\\) + \\(\\frac{3 x^{2}}{2}\\) + 3 log x – \\(\\frac{1}{2 x^{2}}\\) + C<\/p>\n

Question 43.
\nFind \\(\\int \\sqrt{1+\\sin 2 x}\\) dx on R.
\nSolution:
\n1 + sin 2x = sin2<\/sup> x + cos2<\/sup> x + 2 sin x . cos x
\n= (sin x + cos x)2<\/sup>
\n\\(\\sqrt{1+\\sin 2 x}\\) = sin x + cos x
\nIf 2n\u03c0 – \\(\\frac{\\pi}{4}\\) \u2264 x \u2264 2n\u03c0 + \\(\\frac{3\\pi}{4}\\)
\n= -(sin x + cos x). otherwise
\nIf 2n\u03c0 – \\(\\frac{\\pi}{4}\\) \u2264 x \u2264 2n\u03c0 + \\(\\frac{3\\pi}{4}\\), then
\n\\(\\int \\sqrt{1+\\sin 2 x}\\) dx = \\(\\int\\) (sin x + cos x)dx
\n= -cos x + sin x + C
\nIf 2n\u03c0 + \\(\\frac{3\\pi}{4}\\) \u2264 x \u2264 2n\u03c0 + \\(\\frac{7\\pi}{4}\\)
\n\\(\\int \\sqrt{1+\\sin 2 x}\\)
\n= \\(\\int\\) -(sin x + cos x)dx
\n= –\\(\\int\\) sin x dx – \\(\\int\\) cos x dx
\n= cos x – sin x + c<\/p>\n

Question 44.
\nEvaluate \\(\\int \\frac{2 x^{3}-3 x+5}{2 x^{2}}\\) dx for x > 0 and verify the result by differentiation.
\nSolution:
\n\"Inter
\nand it is the given expression and hence the result is correct.<\/p>\n

\"Inter<\/p>\n

Question 45.
\nEvaluate \\(\\int \\frac{x^{5}}{1+x^{12}}\\) dx on R.
\nSolution:
\nWe define f : R \u2192 R by f(t) = \\(\\frac{1}{1+t^{2}}\\)
\ng : R \u2192 R by g(x) = x6<\/sup>
\nThen g'(x) = 6x5<\/sup>
\nDefine F : R \u2192 R by F(t) = tan-1<\/sup> t
\nF is the primitive of f
\n\\(\\int \\frac{x^{5}}{1+x^{12}}\\) dx = \\(\\frac{1}{6}\\) \\(\\int\\) f(g(x)) g'(x) dx
\n= \\(\\frac{1}{6}\\) (F(t) + C)t=g(x)<\/sub>
\n= \\(\\frac{1}{6}\\) [tan-1<\/sup> t + C]t=x6<\/sup><\/sub>
\n= \\(\\frac{1}{6}\\) tan-1<\/sup> x6<\/sup> + C<\/p>\n

Question 46.
\nEvaluate \\(\\int\\) cos3<\/sup> x sin x dx on R.
\nSolution:
\nWe define : f = R \u2192 R by f(x) = cosx
\n\u2234 f'(x) = – sin x
\n\\(\\int\\) cos3<\/sup> x sin x dx = \\(\\int\\) (f(x))3<\/sup> [-f'(x)] dx
\n= \\(\\frac{-[f(x)]^{4}}{4}\\) + C
\n= \\(\\frac{-\\cos ^{4} x}{4}\\) + C<\/p>\n

Question 47.
\nFind \\(\\int\\) (1 – \\(\\frac{1}{x^{2}}\\)) e(x + \\(\\frac{1}{x}\\))<\/sup> dx on I where I = (0, \u221e)
\nSolution:
\nLet J = I = (0, \u221e)
\nDefine f : I \u2192 R by f(t) = et<\/sup> and g : J \u2192 R by g(x) = x + \\(\\frac{1}{x}\\)
\nThen g(J) \u2282 I, g'(x) = 1 – \\(\\frac{1}{x^{2}}\\)
\n\\(\\int\\) (1 – \\(\\frac{1}{x^{2}}\\)) e(x + \\(\\frac{1}{x}\\))<\/sup> dx = \\(\\int\\) f (g(x)) g'(x) dx
\n= \\(\\int\\) [f(t)dt]t = g(x)<\/sub>
\n= [\\(\\int\\)et<\/sup> dt]t = g(x)<\/sub>
\n= [et<\/sup> + c]t=x+\\(\\frac{1}{x}\\)<\/sub>
\n= e(x+\\(\\frac{1}{x}\\))<\/sup> + C<\/p>\n

\"Inter<\/p>\n

Question 48.
\nEvaluate \\(\\int \\frac{1}{\\sqrt{\\sin ^{-1} x} \\sqrt{1-x^{2}}}\\) dx on I = (0, 1).
\nSolution:
\nWe define f : I \u2192 R by f(x) = sin-1<\/sup>x
\nf'(x) = \\(\\frac{1}{\\sqrt{1-x^{2}}}\\)
\n\\(\\int \\frac{1}{\\sqrt{\\sin ^{-1} x} \\sqrt{1-x^{2}}}\\) dx = \\(\\int \\frac{f^{\\prime}(x)}{\\sqrt{f(x)}} d x\\)
\n= 2\\(\\sqrt{f(x)}\\) + C
\n= 2 \\(\\sqrt{\\sin ^{-1} x}\\) + C<\/p>\n

Question 49.
\nEvaluate \\(\\int \\frac{\\sin ^{4} x}{\\cos ^{6} x}\\) dx, x \u2208 I \u2282 R \\ {\\(\\frac{(2 n+1) \\pi}{2}\\) : n \u2208 z}
\nSolution:
\n\\(\\int \\frac{\\sin ^{4} \\dot{x}}{\\cos ^{6} x}\\) dx = \\(\\int\\) tan4<\/sup> x . sec2<\/sup> x dx
\nWe define f : I \u2192 R by f(x)
\nf(x) = tan x, then f'(x) = sec2<\/sup>x
\n\\(\\int \\frac{\\sin ^{4} \\dot{x}}{\\cos ^{6} x}\\) dx = \\(\\int\\) [f(x)]4<\/sup> . f'(x) dx = \\(\\frac{[f(x)]^{5}}{5}\\) + C.
\n= \\(\\frac{1}{5}\\) tan5<\/sup>x + C.<\/p>\n

Question 50.
\nEvaluate \\(\\int\\) sin2<\/sup> x dx on R.
\nSolution:
\n\\(\\int\\) sin2<\/sup>x dx = \\(\\int \\frac{(1-\\cos 2 x)}{2}\\) dx
\n= \\(\\frac{1}{2}\\) \\(\\int\\) dx – \\(\\frac{1}{2}\\) \\(\\int\\) cos 2x dx
\n= \\(\\frac{1}{2}\\) x – \\(\\frac{1}{4}\\) sin 2x + C.
\n(since \\(\\int\\) cos 2x dx = \\(\\frac{1}{2}\\) sin 2x + C)<\/p>\n

Question 51.
\nEvaluate \\(\\int \\frac{1}{a \\sin x+b \\cos x}\\) dx where a, b \u2208 R and a2<\/sup> + b2<\/sup> \u2260 0 on R.
\nSolution:
\nWe can find real numbers r and \u03b8 such that
\na = r cos \u03b8, b = r sin \u03b8
\nThen r = \\(\\sqrt{a^{2}+b^{2}}\\), cos \u03b8 = \\(\\frac{a}{r}\\) and sin \u03b8 = \\(\\frac{b}{r}\\)
\na sin x + b cos x = r . cos \u03b8 sin x + r sin \u03b8 cos x
\n= r[cos \u03b8 sin x + sin \u03b8 cos x]
\n= r sin (x + \u03b8)
\n\\(
\n= [latex]\\frac{1}{r}\\) (cosec (x + \u03b8) dx
\n= \\(\\frac{1}{r}\\) log |tan \\(\\frac{1}{2}\\)(x + \u03b8)| + C
\n= \\(\\frac{1}{\\sqrt{a^{2}+b^{2}}} \\log \\left|\\tan \\frac{1}{2}(\\tilde{x}+\\theta)\\right|+c\\)
\nFor all x \u2208 I where I is an interval disjoint with {n\u03c0 – \u03b8 : n \u2208 z}.<\/p>\n

\"Inter<\/p>\n

Question 52.
\nFind \\(\\int \\frac{x^{2}}{\\sqrt{x+5}}\\) dx on (-5, \u221e)
\nSolution:
\nPut t = x + 5 so that t > 0 on (-5, \u221e)
\ndx = dt and x = t – 5
\n\"Inter<\/p>\n

Question 53.
\nFind \\(\\int \\frac{x}{\\sqrt{1-x}}\\) dx, x \u2208 I = (0, 1)
\nSolution:
\nWe define f : I \u2192 R by f(x) = \\(\\frac{x}{\\sqrt{1-x}}\\)
\nLet J = (0, \\(\\frac{\\pi}{2}\\))
\nDefine \u03a6 : J \u2192 I by \u03a6(\u03b8) = sin2<\/sup> \u03b8
\nThen \u03a6 is a bijective mapping from J to I Further \u03a6 and \u03a6-1<\/sup> are differentiable on their respective domains.
\nput x = \u03a6(\u03b8) = sin2<\/sup>\u03b8
\ndx = 2 sin \u03b8 . cos \u03b8 d\u03b8
\n\"Inter<\/p>\n

Question 54.
\nEvaluate \\(\\int \\frac{d x}{(x+5) \\sqrt{x+4}}\\) on (-4, \u221e)
\nSolution:
\nLet I = (-4, \u221e)
\nDefine f on I as f(x) = \\(\\frac{d x}{(x+5) \\sqrt{x+4}}\\)
\nLet J = (0, \u221e)
\nWe define \u03a6 : J \u2192 I by \u03a6(t) = t2<\/sup> – 4
\nThen \u03a6 is differentiable and is a bijection
\nPut x = \u03a6(t) = t2<\/sup> – 4
\nthen t = \\(\\sqrt{x+4}\\) \u21d2 dx = 2t dt
\nThus \\(\\int \\frac{d x}{(x+5) \\sqrt{x+4}}\\) = \\(\\int \\frac{2 t d t}{\\left(t^{2}+1\\right) t}\\)
\n= \\(\\int \\frac{2 d t}{t^{2}+1}\\)
\n= 2 tan-1<\/sup> t + C
\n= 2 tan-1<\/sup> (\\(\\sqrt{x+4}\\)) + C<\/p>\n

\"Inter<\/p>\n

Question 55.
\nEvaluate \\(\\int \\frac{d x}{\\sqrt{4-9 x^{2}}}\\) on I = (-\\(\\frac{2}{3}\\), \\(\\frac{2}{3}\\))
\nSolution:
\n\"Inter<\/p>\n

Question 56.
\nEvaluate \\(\\int \\frac{1}{a^{2}-x^{2}}\\) dx for x \u2208 I = (-a, a)
\nSolution:
\n\"Inter<\/p>\n

Question 57.
\nEvaluate \\(\\int \\frac{1}{1+4 x^{2}}\\) dx on R.
\nSolution:
\n\\(\\int \\frac{1}{1+4 x^{2}}\\) dx = \\(\\int \\frac{d x}{4\\left(\\left(\\frac{1}{2}\\right)^{2}+x^{2}\\right)}\\)
\n= \\(\\frac{1}{4}\\) (2 tan-1<\/sup> 2x) + C
\n= \\(\\frac{1}{2}\\) tan-1<\/sup> 2x + C<\/p>\n

Question 58.
\nFind \\(\\int \\frac{1}{\\sqrt{4-x^{2}}}\\) dx on (-2, 2).
\nSolution:
\n\\(\\int \\frac{1}{\\sqrt{4-x^{2}}}\\) dx = \\(\\int \\frac{1}{\\sqrt{2^{2}-x^{2}}}\\) dx = sin-1<\/sup>(\\(\\frac{x}{2}\\) dx) + C<\/p>\n

Question 59.
\nEvaluate \\(\\int \\sqrt{4 x^{2}+9}\\) dx on R.
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 60.
\nEvaluate \\(\\int \\sqrt{9 x^{2}-25} d x\\) on [\\(\\frac{5}{3}\\), \u221e)
\nSolution:
\n\"Inter<\/p>\n

Question 61.
\nEvaluate \\(\\int \\sqrt{16-25 x^{2}}\\) dx on (\\(\\frac{-4}{5}\\), \\(\\frac{4}{5}\\))
\nSolution:
\n\"Inter<\/p>\n

Question 62.
\nEvaluate \\(\\int\\) x sin-1<\/sup>x dx on (-1, 1).
\nSolution:
\nLet u(x) = sin-1<\/sup> x and v(x) = \\(\\frac{x^{2}}{2}\\) so that
\nv'(x) = x
\n\u2234 u(x) v'(x) = x sin-1<\/sup>x
\nEven though the domain of u is (-1, 1) the function u \u00a1s differentiable only on (-1, 1).
\nFrom the same formula, we have
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 63.
\nEvaluate \\(\\int\\) x2<\/sup> cos x dx.
\nSolution:
\nLet us take u(x) = x2<\/sup>, v(x) = sin x
\nso that v'(x) = cos x
\nu(x) v'(x) = x2<\/sup> cosx
\nBy using the formula for integration by parts, we have
\n\\(\\int\\) x2<\/sup> cos x dx = x2<\/sup> sin x – \\(\\int\\) sin x (x2<\/sup>)’ dx
\n= x2<\/sup> sin x – 2 \\(\\int\\) x sin x dx + C.
\nAgain, by applying the formula for integration by parts to
\n\\(\\int\\) x sin x dx, we get
\n\\(\\int\\) x. sin x dx = -x cos x – \\(\\int\\) (-cos x) dx
\n= -x cos x + sin x + C2<\/sub>
\n\\(\\int\\) x2<\/sup> cos x dx = x2<\/sup> sin x – 2(sin x – x cos x) + C
\n= x2<\/sup> sin x – 2 sin x + 2x cosx + C
\n= (x2<\/sup> – 2) sin x + 2x cos x + C
\nIn evaluating certain integrals by using the formula for integration by parts,, twice or more than twice, we come across the given integral with change of sign. This enables us to evaluate the given integral.<\/p>\n

Question 64.
\nEvaluate \\(\\int\\) ex<\/sup> sin x dx on R.
\nSolution:
\nLet A = ex<\/sup> sin x dx on R
\nA = \\(\\int\\) ex<\/sup> . sin x dx = \\(\\int\\) ex<\/sup> (-cos x)’ dx.
\n= ex<\/sup> (-cos x) – \\(\\int\\) (-cos x) (ex<\/sup>)’ dx
\n= – ex<\/sup> cos x + \\(\\int\\) ex<\/sup> cos x dx + C1<\/sub> ………(1)
\n\\(\\int\\) ex<\/sup> cos x dx = ex<\/sup>. sin x – \\(\\int\\) ex<\/sup> . sin x dx
\n= ex<\/sup> sinx – A …………….. (2)
\nFrom (1) and (2)
\nA = – ex<\/sup> cos x + ex<\/sup> sin x – A + C1<\/sub>
\n2A = ex<\/sup> (sin x – cos x) + C1<\/sub>
\nA = \\(\\frac{1}{2}\\) ex<\/sup> (sin x – cos x) + C where
\nC = \\(\\frac{C_{1}}{2}\\)
\ni.e., \\(\\int\\) ex<\/sup> sin x dx = \\(\\frac{1}{2}\\) ex<\/sup> (sin x – cos x) + C.<\/p>\n

\"Inter<\/p>\n

Question 65.
\nFind \\(\\int\\) eax<\/sup> cos (bx + c) dx on R, where a. b, c are real numbers and b \u2260 0.
\nSolution:
\nLet A = \\(\\int\\) eax<\/sup> cos (bx + c)dx
\nThen from the formula for integration by parts
\nA = eax<\/sup> [latex]\\frac{\\sin (b x+c)}{b}[\/latex] – \\(\\int\\) a eax<\/sup> [latex]\\frac{\\sin (b x+c)}{b}[\/latex] dx
\n= \\(\\frac{1}{b}\\) eax<\/sup> sin(bx + c) – \\(\\frac{a}{b}\\) \\(\\int\\) eax<\/sup> . sin(bx + c) dx
\n\"Inter
\nBy taking c = 0, we get
\n\\(\\int\\) eax<\/sup> . cos bx dx = \\(\\frac{e^{a x}}{a^{2}+b^{2}}\\) [a cos bx + b sin bx] + K<\/p>\n

Question 66.
\nEvaluate \\(\\int \\tan ^{-1} \\sqrt{\\frac{1-x}{1+x}}\\) dx, on (-1, 1)
\nSolution:
\nPut x = cos \u03b8, \u03b8 \u2208 (0, \u03c0) dx = -sin \u03b8 . d\u03b8
\n\"Inter<\/p>\n

Question 67.
\nEvaluate \\(\\int e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right) d x\\) on I \u2282 R \\ {2n\u03c0 : n \u2208 z}.
\nSolution:
\n\"Inter
\n= -ex<\/sup> . cot \\(\\frac{x}{2}\\) + C<\/p>\n

\"Inter<\/p>\n

Question 68.
\nEvaluate \\(\\int \\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right) d x\\) on I \u2282 R \\ {-1, 1}
\nSolution:
\nLet x = tan \u03b8 \u21d2 dx = sec2<\/sup> \u03b8 d\u03b8
\n\\(\\frac{2 x}{1-x^{2}}=\\frac{2 \\tan \\theta}{1-\\tan ^{2} \\theta}\\) = tan 2\u03b8
\ntan-1<\/sup> \\(\\left(\\frac{2 x}{1-x^{2}}\\right)\\) = tan-1<\/sup> (tan 2\u03b8 ) = 2\u03b8 + n\u03c0
\nWhere n = 0 if |x| < 1 = -1 if x > 1
\n= 1 if x < -1
\nWe have d\u03b8 = \\(\\frac{1}{1+x^{2}}\\) dx and
\n1 + x2<\/sup> = 1 + tan2<\/sup> \u03b8 = sec2<\/sup>\u03b8
\n\u2234 \\(\\int \\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right) d x\\)
\n= \\(\\int\\left(\\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right)\\right)\\left(1+x^{2}\\right) \\frac{1}{1+x^{2}} d x\\)
\n= \\(\\int\\) (2\u03b8 + n\u03c0) \\(\\int\\) sec2<\/sup>\u03b8 d\u03b8
\n= 2 \\(\\int\\) \u03b8 sec2<\/sup> \u03b8 d\u03b8 + n\u03c0 ) \\(\\int\\) sec2<\/sup> \u03b8 d\u03b8 + c
\n= 2 (\u03b8 tan \u03b8 – \\(\\int\\) tan \u03b8 d\u03b8) n\u03c0 tan \u03b8 + c
\n= 2 (\u03b8 tan \u03b8 + log |cos \u03b8| + n\u03c0 tan \u03b8 + c
\n= (2\u03b8 + n\u03c0) tan \u03b8 + 2 log cos \u03b8 + c
\n= (2\u03b8 + n\u03c0) tan \u03b8 + log cos2<\/sup> \u03b8 + c
\n= (2\u03b8 + n\u03c0) tan \u03b8 + log sec2<\/sup> \u03b8 + c
\n= x tan-1<\/sup> \\(\\left(\\frac{2 x}{1-x^{2}}\\right)\\) – log (1 + x2<\/sup>) + c<\/p>\n

Question 69.
\nFind \\(\\int x^{2} \\cdot \\frac{\\exp \\left(m \\sin ^{-1} x\\right)}{\\sqrt{1-x^{2}}}\\)dx on (-1, 1) where m is a real number. (Here for y \u2208 R, exp. {y} stands for ey<\/sup>).
\nSolution:
\nLet t = sin-1<\/sup>x, then
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 70.
\nEvaluate \\(\\int \\frac{d x}{4 x^{2}-4 x-7}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 71.
\nFind \\(\\int \\frac{d x}{5-2 x^{2}+4 x}\\)
\nSolution:
\n5 – 2x2<\/sup> + 4x = -2 (x2<\/sup> – 2x – \\(\\frac{5}{2}\\))
\n= -2 ((x – 1)2<\/sup> – \\(\\frac{5}{2}\\) – 1)
\n= -2 ((x – 1)2<\/sup> – \\(\\left(\\sqrt{\\frac{7}{2}}\\right)^{2}\\))
\n\\(\\int \\frac{d x}{5-2 x^{2}+4 x}\\)
\n= \\(-\\frac{1}{2} \\int \\frac{1}{\\left((x-1)^{2}-\\sqrt{\\frac{7}{2}}\\right)^{2}} d x+C\\)
\n\"Inter<\/p>\n

Question 72.
\nEvaluate \\(\\int \\frac{d x}{x^{2}+x+1}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 73.
\nEvaluate \\(\\int \\frac{d x}{\\sqrt{x^{2}+2 x+10}}\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 74.
\nEvaluate \\(\\int \\frac{d x}{\\sqrt{1+x-x^{2}}}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 75.
\nEvaluate \\(\\int \\sqrt{3+8 x-3 x^{2}} d x\\)
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 76.
\nEvaluate \\(\\int \\frac{x+1}{x^{2}+3 x+12}\\) dx.
\nSolution:
\nWe write x + 1 = A(2x + 3) + B
\nEquating the co-efficients of x; we get 1 = 2A.
\nA = \\(\\frac{1}{2}\\)
\nEquating the constants 3A + B = 1
\nB = 1 – 3A = 1 – \\(\\frac{3}{2}\\) = –\\(\\frac{1}{2}\\)
\nx + 1 = \\(\\frac{1}{2}\\) (2x + 3) – \\(\\frac{1}{2}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 77.
\nEvaluate \\(\\int(3 x-2) \\sqrt{2 x^{2}-x+1} d x\\)
\nSolution:
\nLet (3x – 2) = A(4x – 1) + B
\nEquating the co-efficients of x, we get 3 = 4A
\nA = \\(\\frac{3}{4}\\)
\nEquating the constants -2 = -A + B
\nB = -2 + A = -2 + \\(\\frac{3}{4}\\)
\n= \\(\\frac{-5}{4}\\)
\n\"Inter
\n\"Inter<\/p>\n

Question 78.
\nEvaluate \\(\\int \\frac{2 x+5}{\\sqrt{x^{2}-2 x+10}}\\)dx. [T.S. Mar. 15]
\nSolution:
\nWe write
\n2x + 5 = A \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\) (x2<\/sup> – 2x + 10) + B
\n= A (2x – 2) + B
\nOn comparing the coefficients of the like powers of x on both sides of the above equation, we get A = 1 and B = 7.
\nThus 2x + 5 = (2x – 2) + 7
\nHence \\(\\int \\frac{2 x+5}{\\sqrt{x^{2}-2 x+10}}\\)dx
\n= \\(\\int \\frac{2 x-2}{\\sqrt{x^{2}-2 x+10}} d x+7 \\int \\frac{d x}{\\sqrt{x^{2}-2 x+10}}+C\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 79.
\nEvaluate \\(\\int \\frac{d x}{(x+5) \\sqrt{x+4}}\\)
\nSolution:
\nPut t = \\(\\sqrt{x+4}\\)
\ndt = \\(\\frac{1}{2 \\sqrt{x+4}}\\) dx
\nWe have t2<\/sup> = x + 4
\nx + 5 = t2<\/sup> + 1
\n\\(\\int \\frac{d x}{(x+5)(\\sqrt{x+4})}=\\int \\frac{2}{t^{2}+1} d t\\)
\n= 2 tan-1<\/sup> t + C
\n= 2 tan-1<\/sup> (\\(\\sqrt{x+4}\\)) + C.<\/p>\n

Question 80.
\nEvaluate \\(\\int \\frac{d x}{5+4 \\cos x}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 81.
\nFind \\(\\int \\frac{d x}{3 \\cos x+4 \\sin x+6}\\) [Mar. 13]
\nSolution:
\nt = tan \\(\\frac{x}{2}\\) \u21d2 dx = \\(\\frac{2 d t}{1+t^{2}}\\)
\n\"Inter
\n\"Inter<\/p>\n

Question 82.
\nFind \\(\\int \\frac{d x}{d+e \\tan x}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 83.
\nEvaluate \\(\\int \\frac{\\sin x}{d \\cos x+e \\sin x} d x\\) and \\(\\int \\frac{\\cos x}{d \\cos x+e \\sin x} d x\\).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 84.
\nEvaluate \\(\\int \\frac{\\cos x+3 \\sin x+7}{\\cos x+\\sin x+1} d x\\)
\nSolution:
\nLet cos x + 3 sin x + 7 = A(cos x + sin x + 1)’ + B(cos x + sin x + 1) + C
\nComparing the coefficients
\nA + B = 1, A – B = 3, B + C = 7
\nA = -1, B = 2, C = 5
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 85.
\nFind \\(\\int \\frac{x^{3}-2 x+3}{x^{2}+x-2}\\) dx.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 86.
\nFind \\(\\int \\frac{d x}{x^{2}-81}\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 87.
\nFind \\(\\int \\frac{2 x^{2}-5 x+1}{x^{2}\\left(x^{2}-1\\right)}\\) dx.
\nSolution:
\nLet \\(\\frac{2 x^{2}-5 x+1}{x^{2}\\left(x^{2}-1\\right)}=\\frac{A}{x}+\\frac{B}{x^{2}}+\\frac{C}{x-1}+\\frac{D}{x+1}\\)
\n2x2<\/sup> – 5x + 1 = Ax(x2<\/sup> – 1) + B(x2<\/sup> – 1) + Cx2<\/sup> (x + 1) + Dx2<\/sup> (x – 1)
\nx = 1 \u21d2 2 – 5 + 1 = C (1 + 1) \u21d2 2C = -2
\n\u21d2 C = -1
\nx = -1 \u21d2 2 + 5 + 1 = D (-1 -1)
\n\u21d2 8 = -2B \u21d2 D = -4
\nx = 0 \u21d2 1 = B(-1) \u21d2 B = -1
\nEquating the coefficients of x3<\/sup>
\n0 = A + C + D \u21d2 A = -C – D = 1 + 4 = 5
\n\"Inter<\/p>\n

Question 88.
\nFind \\(\\int \\frac{3 x-5}{x\\left(x^{2}+2 x+4\\right)}\\) dx.
\nSolution:
\n\\(\\frac{3 x-5}{x\\left(x^{2}+2 x+4\\right)}=\\frac{A}{x}+\\frac{B x+C}{x^{2}+2 x+4}\\)
\n3x – 5 = A(x2<\/sup> + 2x + 4) + (Bx + C) . x
\nx = 0 \u21d2 -5 = 4 A \u21d2 A = –\\(\\frac{5}{4}\\)
\nEquating the coefficients of x2<\/sup>
\nA + B = 0 \u21d2 B = -A = \\(\\frac{5}{4}\\)
\nEquating the coefficient of x
\n3 = 2 A + C
\nC = 3 – 2 A = 3 + 2 . \\(\\frac{5}{4}\\) = \\(\\frac{11}{2}\\)
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 89.
\nFind \\(\\int \\frac{2 x+1}{x\\left(x^{2}+4\\right)^{2}} d x\\)
\nSolution:
\nLet \\(\\frac{2 x+1}{x\\left(x^{2}+4\\right)^{2}}=\\frac{A}{x}+\\frac{B x+C}{x^{2}+4}+\\frac{D x+E}{\\left(x^{2}+4\\right)^{2}}\\)
\n2x + 1 = A (x2<\/sup> + 4)2<\/sup> + (Bx + C) + x (x2<\/sup> + 4) + (Dx + E)x
\nEquating the coefficients of like power of x, we obtain
\nA + B = 0, C = 0, 8A + 4B + D = 0,
\n4C + E = 2, A = \\(\\frac{1}{16}\\)
\nSolving these equation, we obtain
\n\"Inter
\n\"Inter<\/p>\n

Question 90.
\nEvaluate \\(\\int\\) x3<\/sup> . e5x<\/sup> dx.
\nSolution:
\nWe take a = 5, use the reduction formula
\n\"Inter
\n\"Inter<\/p>\n

Question 91.
\nEvaluate \\(\\int\\) sin4<\/sup> x dx. [Mar. 14]
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 92.
\nEvaluate \\(\\int\\) tan6<\/sup> x dx.
\nSolution:
\n\"Inter<\/p>\n

Question 93.
\n\\(\\int\\) sec5<\/sup> x dx.
\nSolution:
\nReduction formula is
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Integration Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2B Integration Important Questions Question 1. [May 11] Solution: Question 2. sec2x cosec2x dx on I \u2282 R \\ ({n\u03c0 : n \u2208 Z} \u222a {(2n … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22137"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22137"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22137\/revisions"}],"predecessor-version":[{"id":22139,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22137\/revisions\/22139"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22137"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22137"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22137"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}