{"id":9972,"date":"2024-02-22T09:31:17","date_gmt":"2024-02-22T04:01:17","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9972"},"modified":"2024-02-24T18:00:11","modified_gmt":"2024-02-24T12:30:11","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6e","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6e\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(e)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 6 Integration Ex 6(e) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(e)<\/h2>\n

I. Evaluate the following integrals.<\/span><\/p>\n

Question 1.
\n\\(\\int \\frac{x-1}{(x-2)(x-3)} d x\\)
\nSolution:
\n\\(\\int \\frac{x-1}{(x-2)(x-3)} d x\\)
\n\"TS
\n= x + 2 log|x – 3| – x – log|x – 2| + c
\n= 2 log|x – 3| – log|x – 2| + c
\nAlter: Let \\(\\frac{x-1}{(x-2)(x-3)}=\\frac{A}{x-2}+\\frac{B}{x-3}\\) (Partial fractions method)
\n\u2234 x – 1 = A(x – 3) + B(x – 2)
\nComparing coefficients of x and constant terms on both sides
\nA + B = 1 ………(1) and
\n-3A – 2B = -1
\n\u21d2 3A + 2B = 1 ………(2)
\nFrom (1), 2A + 2B = 2
\nSolving A = -1 and B = 2
\n\"TS
\n= -log|x – 2| + 2 log|x – 3| + c
\n= 2 log|x – 3| – log|x – 2| + c<\/p>\n

\"TS<\/p>\n

Question 2.
\n\\(\\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{A}{x+2}+\\frac{C}{(x+2)^2}\\)
\n\u2234 x2<\/sup> = A(x + 2)2<\/sup> + B(x + 1)(x + 2) + C(x + 1)
\nPut x = -1, then A = 1
\nComparing the coefficient of x2<\/sup>,
\nA + B = 1 \u21d2 B = 0
\nPut x = -2, and we get
\n4 = C(-1)
\n\u21d2 C = -4
\n\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{x+3}{(x-1)\\left(x^2+1\\right)} d x\\)
\nSolution:
\nLet \\(\\frac{x+3}{(x-1)\\left(x^2+1\\right)}=\\frac{A}{x-1}+\\frac{B x+C}{x^2+1}\\)
\n\u2234 x + 3 = A(x2<\/sup> + 1) + (Bx + C)(x – 1)
\nPut x = 1, then 4 = 2A
\n\u21d2 A = 2
\nComparing the coefficient of x2<\/sup>,
\nA + B = 0
\n\u21d2 B = -A = -2
\nComparing the coefficient of constant terms
\nA – C = 3
\n\u21d2 C = A – 3 = 2 – 3 = -1
\n\u2234 x + 3 = 2(x2<\/sup> + 1) + (-2x – 1)(x – 1) = 2(x2<\/sup> + 1) – (2x + 1)(x – 1)
\n\"TS<\/p>\n

Question 4.
\n\\(\\int \\frac{d x}{\\left(x^2+a^2\\right)\\left(x^2+b^2\\right)}\\)
\nSolution:
\n\"TS<\/p>\n

Question 5.
\n\\(\\int \\frac{d x}{e^x+e^{2 x}}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 6.
\n\\(\\int \\frac{d x}{(x+1)(x+2)}\\) (Mar. ’12; May ’11)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 7.
\n\\(\\int \\frac{1}{e^x-1} d x\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 8.
\n\\(\\int \\frac{1}{(1-x)\\left(4+x^2\\right)} d x\\)
\nSolution:
\nLet \\(\\frac{1}{(1-x)\\left(4+x^2\\right)}=\\frac{A}{1-x}+\\frac{B x+C}{x^2+4}\\)
\n\u2234 1 = A(x2<\/sup> + 4) + (Bx + C)(1 – x)
\nWhen x – 1 then 5A = 1 \u21d2 A = \\(\\frac{1}{5}\\)
\nThe coefficient of x2<\/sup> on both sides gives
\nA – B = 0 \u21d2 B = \\(\\frac{1}{5}\\)
\nComparing constant terms,
\n\"TS<\/p>\n

Question 9.
\n\\(\\int \\frac{2 x+3}{x^3+x^2-2 x} d x\\)
\nSolution:
\nx3<\/sup> + x2<\/sup> – 2x = x(x2<\/sup> + x – 2) = x(x – 1)(x + 2)
\n\"TS
\n\u2234 2x + 3 = A(x – 1)(x + 2) + Bx(x + 2) + Cx(x – 1)
\nPut x = 1 we get
\n5 = 3B \u21d2 B = \\(\\frac{5}{3}\\)
\nPut x = -2 on both sides
\n-4 + 3 = C(-2) (-2 – 1)
\n\u21d2 -1 = 6C
\n\u21d2 C = \\(-\\frac{1}{6}\\)
\nCoefficient of x2 on both sides
\nA + B + C = 0
\n\"TS<\/p>\n

II. Evaluate the following integrals.<\/span><\/p>\n

Question 1.
\n\\(\\int \\frac{d x}{6 x^2-5 x+1}\\)
\nSolution:
\n6x2<\/sup> – 5x + 1 = 6x2<\/sup> – 3x – 2x + 1
\n= 3x(2x – 1) – 1(2x – 1)
\n= (3x – 1)(2x – 1)
\n\"TS<\/p>\n

Question 2.
\n\\(\\int \\frac{d x}{x(x+1)(x+2)}\\)
\nSolution:
\nLet \\(\\frac{1}{x(x+1)(x+2)}=\\frac{A}{x}+\\frac{B}{x+1}+\\frac{C}{x+2}\\)
\n\u2234 1 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)
\nPut x = -1 then -B = 1 \u21d2 B = -1
\nCoefficient of x2 both sides
\nA + B + C = 0 and put x = – 2 then
\nC(-2)(-1) = 1
\n\u21d2 C = \\(\\frac{1}{2}\\)
\n\u2234 A = -B – C = 1 – \\(\\frac{1}{2}\\) = \\(\\frac{1}{2}\\)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{3 x-2}{(x-1)(x+2)(x-3)} d x\\)
\nSolution:
\nLet \\(\\frac{3 x-2}{(x-1)(x+2)(x-3)}=\\frac{A}{x-1}+\\frac{B}{x+2}\\) + \\(\\frac{C}{x-3}\\)
\n\u2234 3x – 2 = A(x + 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x + 2)
\nPut x = 1, then 1 = A(3)(-2) \u21d2 A = \\(-\\frac{1}{6}\\)
\nPut x = -2, then -8 = B(-3)(-5) \u21d2 B = \\(-\\frac{8}{15}\\)
\nPut x = 3, then 7 = C(2)(5) \u21d2 C = \\(\\frac{7}{10}\\)
\n\"TS<\/p>\n

Question 4.
\n\\(\\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\u2234 7x – 4 = A(x – 1)(x + 2) + B(x + 2) + C(x – 1)2<\/sup>
\nPut x = 1 both sides 3 = 3B \u21d2 B = 1
\nPut x = -2 both sides -18 = 9C \u21d2 C = -2
\nComparing the coefficient of x2<\/sup> on both sides
\nA + C = 0
\n\u21d2 A = -C = 2
\n\"TS<\/p>\n

III. Evaluate the following integrals.<\/span><\/p>\n

Question 1.
\n\\(\\int \\frac{1}{(x-a)(x-b)(x-c)} d x\\)
\nSolution:
\nLet \\(\\frac{1}{(x-a)(x-b)(x-c)}\\) = \\(\\frac{A}{x-a}+\\frac{B}{x-b}+\\frac{C}{x-c}\\)
\nThen 1 = A(x – b)(x – c) + B(x – a)(x – c) + C(x – a)(x – b)
\nPut x = a both sides we get 1 = A(a – b)(a – c)
\n\u21d2 A =\\(\\frac{1}{(a-b)(a-c)}\\)
\nPut x = b both sides we get 1 = B(b – a)(b – c)
\n\u21d2 B = \\(\\frac{1}{(b-a)(b-c)}\\)
\nPut x = c both sides we get 1 = C (c – a)(c – b)
\n\u21d2 C = \\(\\frac{1}{(c-a)(c-b)}\\)
\n\"TS<\/p>\n

Question 2.
\n\\(\\int \\frac{2 x+3}{(x+3)\\left(x^2+4\\right)} d x\\)
\nSolution:
\nLet \\(\\frac{2 x+3}{(x+3)\\left(x^2+4\\right)}=\\frac{A}{x+3}+\\frac{B x+C}{x^2+4}\\) ……(1)
\n\u2234 2x + 3 = A(x2<\/sup> + 4) + (Bx + c)(x + 3)
\nPut x = -3 both sides -6 + 3 = A(9 + 4)
\n\u21d2 A = \\(-\\frac{3}{13}\\)
\nComparing the coefficient of x2<\/sup> on both sides
\nA + B = 0
\n\u21d2 B = -A = \\(\\frac{3}{13}\\)
\nand comparing constant terms on both sides
\n4A + 3C = 3
\n\u21d2 3C = 3 – 4A
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{2 x^2+x+1}{(x+3)(x-2)^2} d x\\)
\nSolution:
\nLet \\(\\frac{2 x^2+x+1}{(x+3)(x-2)^2}=\\frac{A}{x+3}+\\frac{B}{x-2}\\) + \\(\\frac{C}{(x-2)^2}\\)
\n\u2234 2x2<\/sup> + x + 1 = A(x – 2)2<\/sup> + B(x – 2)(x + 3) + C(x + 3)
\nPut x = 2 on both sides
\n8 + 2 + 1 = C(5)
\n\u21d2 C = \\(\\frac{11}{5}\\)
\nPut x = -3 on both sides
\n18 – 3 + 1 = A(-5)2<\/sup>
\n\u21d2 A = \\(\\frac{16}{25}\\)
\nPut x = 0 on both sides
\n\"TS
\n\"TS<\/p>\n

Question 4.
\n\\(\\int \\frac{d x}{x^3+1}\\)
\nSolution:
\nWe have x3<\/sup> + 1 = (x + 1)(x2<\/sup> – x + 1)
\n\u2234 \\(\\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}\\)
\n\u2234 1 = A(x2<\/sup> – x + 1) + (Bx + C)(x + 1) …….(1)
\nPut x = -1 both sides 1 = A(3)
\n\u21d2 A = \\(\\frac{1}{3}\\)
\nComparing the coefficient of x2<\/sup> on both sides
\nA + B = 0
\n\u21d2 B = \\(-\\frac{1}{3}\\)
\nComparing constant terms we get
\nA + C = 1
\n\u21d2 C = 1 – A
\n= 1 – \\(\\frac{1}{3}\\)
\n= \\(\\frac{2}{3}\\)
\n\u2234 From(1)
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\n\\(\\int \\frac{\\sin x \\cos x}{\\cos ^2 x+3 \\cos x+2} d x\\)
\nSolution:
\nLet cos x = t then sin x dx = -dt
\n\"TS
\n\u2234 t = A(t + 1) + B(t + 2)
\nPut t = -1 then -1 = B(-1 + 2)
\n\u21d2 B = -1
\nand when t = -2 then -2 = A(-1)
\n\u21d2 A = 2
\n\u2234 \\(\\int \\frac{t}{t^2+3 t+2} d t=\\int \\frac{2}{t+2} d t-\\int \\frac{1}{t+1} d t\\)
\n\u2234 From (1)
\n\"TS
\n= log|t + 1| – 2 log|t + 2|
\n= log|cos x + 1| – 2 log|cos x + 2| + c<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(e) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(e) I. Evaluate the following integrals. Question 1. Solution: = x + 2 log|x – 3| – x – log|x – … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9972"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9972"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9972\/revisions"}],"predecessor-version":[{"id":10001,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9972\/revisions\/10001"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9972"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9972"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9972"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}