{"id":9740,"date":"2024-02-24T10:24:16","date_gmt":"2024-02-24T04:54:16","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9740"},"modified":"2024-02-26T09:32:40","modified_gmt":"2024-02-26T04:02:40","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6d","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6d\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(d)"},"content":{"rendered":"

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

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

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

Question 1.
\n\\(\\int \\frac{1}{\\sqrt{2 x-3 x^2+1}} d x\\)
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 2.
\n\\(\\int \\frac{\\sin \\theta}{\\sqrt{2-\\cos ^2 \\theta}} d \\theta\\)
\nSolution:
\nLet cos \u03b8 = t, then sin \u03b8 d\u03b8 = -dt
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{\\cos x}{\\sin ^2 x+4 \\sin x+5} d x\\)
\nSolution:
\nLet sin x = t, then cos x dx = dt
\n\"TS<\/p>\n

Question 4.
\n\\(\\int \\frac{d x}{1+\\cos ^2 x}\\)
\nSolution:
\nDividing by cos2<\/sup>x we get
\n\"TS
\n\"TS<\/p>\n

Question 5.
\n\\(\\int \\frac{d x}{2 \\sin ^2 x+3 \\cos ^2 x}\\)
\nSolution:
\nDividing by cos2<\/sup>x we get
\n\"TS<\/p>\n

Question 6.
\n\\(\\int \\frac{1}{1+\\tan x} d x\\)
\nSolution:
\n\"TS
\nwrite cos x = A(sin x + cos x) + B \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(sin x + cos x)
\n= A(sin x + cos x) + B(cos x – sin x)
\nComparing coefficients of cos x and sin x on both sides
\nA – B = 0 and A + B = 1
\nSolving 2A = 1 \u21d2 A = \\(\\frac{1}{2}\\)
\n\u2234 B = \\(\\frac{1}{2}\\)
\n\u2234 cos x = \\(\\frac{1}{2}\\)(sin x + cos x) + \\(\\frac{1}{2}\\)(cos x – sin x)
\n\"TS
\n= \\(\\frac{1}{2}\\)x + \\(\\frac{1}{2}\\) log|sin x + cos x| + c
\n[\u2235 sin x + cos x = t \u21d2 (cos x – sin x) dx = dt in second integral]<\/p>\n

\"TS<\/p>\n

Question 7.
\n\\(\\int \\frac{1}{1-\\cot x} d x\\)
\nSolution:
\n\"TS
\nLet sin x = A(sin x – cos x) + B \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(sin x – cos x)
\n= A(sin x – cos x) + B(cos x + sin x)
\nComparing coefficients of sin x and cos x on both sides we get
\nA + B = 1 and -A + B = 0
\nsolving B = \\(\\frac{1}{2}\\) and A = \\(\\frac{1}{2}\\)
\n\"TS
\n= \\(\\frac{1}{2}\\)x + \\(\\frac{1}{2}\\) log|sin x – cos x| + c
\n[\u2235 sin x – cos x = t \u21d2 (cos x + sin x) dx = dt in second integral]<\/p>\n

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

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

Question 2.
\n\\(\\int\\left(\\frac{9 \\cos x-\\sin x}{4 \\sin x+5 \\cos x}\\right) d x\\) (New Model Paper)
\nSolution:
\nLet 9 cos x – sin x = A(4 sin x + 5 cos x) + B \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(4 sin x + 5 cos x)
\n\u2234 9 cos x – sin x = A(4 sin x + 5 cos x) + B(4 cos x – 5 sin x) …….(1)
\nComparing coefficients of cos x and sin x on both sides
\n5A + 4B = 9 …….(2)
\nand 4A – 5B = -1 ……….(3)
\nSolving (2) and (3)
\n\"TS
\n= x + log|4 sin x + 5 cos x| + c
\n(\u2235 4 sin x + 5 cos x = t \u21d2 (4 cos x – 5 sin x) dx = dt in second integral)<\/p>\n

Question 3.
\n\\(\\int \\frac{2 \\cos x+3 \\sin x}{4 \\cos x+5 \\sin x} d x\\)
\nSolution:
\nLet 2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B \\(\\frac{d}{d x}\\)(4 cos x + 5 sin x)
\n= A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x) ……..(1)
\nComparing coefficients of cos x and sin x on both sides we get
\n4A + 5B = 2 ………(2)
\nand 5A – 4B = 3 ………(3)
\nSolving (2) and (3) we get
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 4.
\n\\(\\int \\frac{1}{1+\\sin x+\\cos x} d x\\)
\nSolution:
\n\"TS<\/p>\n

Question 5.
\n\\(\\int \\frac{1}{3 x^2+x+1} d x\\)
\nSolution:
\n\"TS<\/p>\n

Question 6.
\n\\(\\int \\frac{d x}{\\sqrt{5-2 x^2+4 x}}\\)
\nSolution:
\nConsider 5 – 2x2<\/sup> + 4x = 5 – (2x2<\/sup> – 4x)
\n\"TS
\n\"TS<\/p>\n

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

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

Question 2.
\n\u222b(6x + 5) \\(\\sqrt{6-2 x^2+x}\\) dx (Mar. ’09)
\nSolution:
\nLet 6x + 5 = A \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(6 – 2x2<\/sup> + x) + B
\n= A(-4x + 1) + B
\nEquating the coefficients of x and constant terms
\n-4A = 6 \u21d2 A = \\(-\\frac{3}{2}\\)
\nand A + B = 5
\n\u21d2 B = 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\n\"TS
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{d x}{4+5 \\sin x}\\)
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 4.
\n\\(\\int \\frac{1}{2-3 \\cos 2 x} d x\\) (June ’10)
\nSolution:
\nLet tan x = t then sec2<\/sup>x dx = dt
\n\"TS
\n\"TS<\/p>\n

Question 5.
\n\u222bx\\(\\sqrt{1+x-x^2}\\) dx (May ’12)
\nSolution:
\nLet x = A \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(1 + x – x2<\/sup>) + B = A(1 – 2x) + B
\nComparing the coefficient of x, constant terms on both sides
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 6.
\n\\(\\int \\frac{d x}{(1+x) \\sqrt{3+2 x-x^2}}\\) (New Model Paper)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 7.
\n\\(\\int \\frac{d x}{4 \\cos x+3 \\sin x}\\)
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\n\\(\\int \\frac{1}{\\sin x+\\sqrt{3} \\cos x} d x\\) (May ’12)
\nSolution:
\n\"TS
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 9.
\n\\(\\int \\frac{d x}{5+4 \\cos 2 x}\\) (Mar. ’11)
\nSolution:
\n\"TS<\/p>\n

Question 10.
\n\\(\\int \\frac{2 \\sin x+3 \\cos x+4}{3 \\sin x+4 \\cos x+5} d x\\) (Mar. ’11)
\nSolution:
\nSince there exist constants in both the numerator and denominator, we determine constants A, B, and C such that
\n2 sin x + 3 cos x + 4 = A \\(\\frac{\\mathrm{d}}{\\mathrm{dx}}\\)(3 sin x + 4 cos x + 5) + B (3 sin x + 4 cos x + 5) + C
\n= A(3 cos x – 4 sin x) + B(3 sin x + 4 cos x + 5) + C …….(1)
\nComparing both sides the coefficients of sin x, cos x, and constants
\n-4A + 3B = 2
\n\u21d2 4A – 3B + 2 = 0 ……..(2)
\n3A + 4B – 3 = 0 ……….(3)
\n5B + C – 4 = 0 ………(4)
\nSolving (2) and (3)
\n\"TS
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 11.
\n\\(\\int \\sqrt{\\frac{5-x}{x-2}} d x\\) on (2, 5).
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 12.
\n\\(\\int \\sqrt{\\frac{1+x}{1-x}} d x\\) on (-1, 1).
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 13.
\n\\(\\int \\frac{d x}{(1-x) \\sqrt{3-2 x-x^2}}\\) on (-1, 3).
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 14.
\n\\(\\int \\frac{d x}{(x+2) \\sqrt{x+1}}\\) on (-1, \u221e).
\nSolution:
\nLet x + 1 = t2<\/sup> then dx = 2t dt
\n\"TS<\/p>\n

Question 15.
\n\\(\\int \\frac{d x}{(2 x+3) \\sqrt{x+2}}\\) on I \u2282 (-2, \u221e) \\ {\\(-\\frac{3}{2}\\)}
\nSolution:
\nLet x + 2 = t2<\/sup> then dx = 2t dt
\nand 2x + 3 = 2(t2<\/sup> – 2) + 3 = 2t2<\/sup> – 1
\n\"TS<\/p>\n

Question 16.
\n\\(\\int \\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}} d x\\) on (0, 1).
\nSolution:
\nPut x = t2<\/sup> then dx = 2t dt
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 17.
\n\\(\\int \\frac{d x}{(x+1) \\sqrt{2 x^2+3 x+1}}\\) on I \u2282 R \\ [-1, \\(-\\frac{1}{2}\\)]
\nSolution:
\nLet x + 1 = \\(\\frac{1}{t}\\)
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 18.
\n\\(\\int \\sqrt{e^x-4} d x\\) on \\(\\left[\\log _e 4, \\infty\\right)\\).
\nSolution:
\nLet ex<\/sup> – 4 = t2<\/sup> then ex dx = 2t dt
\n\"TS
\n\"TS<\/p>\n

Question 19.
\n\\(\\int \\sqrt{1+\\sec x} d x\\) on \\(\\left[\\left(2 n-\\frac{1}{2}\\right) \\pi,\\left(2 n+\\frac{1}{2}\\right) \\pi\\right]\\), n \u2208 Z.
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 20.
\n\\(\\int \\frac{d x}{1+x^4}\\) on R.
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(d) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(d) I. Evaluate the following integrals. Question 1. Solution: Question 2. Solution: Let cos \u03b8 = t, then sin \u03b8 d\u03b8 … 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\/9740"}],"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=9740"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9740\/revisions"}],"predecessor-version":[{"id":9839,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9740\/revisions\/9839"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9740"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9740"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9740"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}