{"id":25428,"date":"2022-09-11T18:46:17","date_gmt":"2022-09-11T13:16:17","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=25428"},"modified":"2022-09-11T18:46:17","modified_gmt":"2022-09-11T13:16:17","slug":"inter-2nd-year-maths-2b-integration-solutions-ex-6d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2b-integration-solutions-ex-6d\/","title":{"rendered":"Inter 2nd Year Maths 2B Integration Solutions Ex 6(d)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Integration Solutions Exercise 6(d) will help students to clear their doubts quickly.<\/p>\n

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

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

Question 1.
\n\u222b\\(\\frac{dx}{\\sqrt{2x-3x^2+1}}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\n\u222b\\(\\frac{\\sin \\theta}{\\sqrt{2-\\cos^2 \\theta}}\\)d\u03b8
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n\u222b\\(\\frac{\\cos x}{\\sin^2 x+4sin x+5}\\)dx
\nSolution:
\nt = sin x \u21d2 dt = cos x dx
\nI = \u222b\\(\\frac{dt}{t^2+4t+5}\\) = \u222b\\(\\frac{dt}{(t+2)^2+1}\\)
\n= tan-1<\/sup>(t + 2) + C
\n= tan-1<\/sup>(sin x + 2) + C<\/p>\n

\"Inter<\/p>\n

Question 4.
\n\u222b\\(\\frac{dx}{1+\\cos^2 x}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 5.
\n\u222b\\(\\frac{dx}{2\\sin^2 x+3\\cos^2 x}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\n\u222b\\(\\frac{1}{1+\\tan x}\\)dx
\nSolution:
\n\"Inter<\/p>\n

Question 7.
\n\u222b\\(\\frac{1}{1-\\cot x}\\)dx
\nSolution:
\n\"Inter<\/p>\n

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

Question 1.
\n\u222b\\(\\sqrt{1+3x-x^2}\\)dx
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\n\u222b\\(\\frac{9\\cos x-\\sin x}{4\\sin x+5\\cos x}\\)dx
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n\u222b\\(\\frac{2\\cos x+3\\sin x}{4\\cos x+5\\sin x}\\)dx
\nSolution:
\nLet 2 cos c + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 cos x)
\nEquating the co-efficient of sin x and cos x,
\nwe get
\n4A + 5B = 2
\n5A – 4B = 3
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 4.
\n\u222b\\(\\frac{1}{1+\\sin x+\\cos x}\\)dx
\nSolution:
\n\"Inter<\/p>\n

Question 5.
\n\u222b\\(\\frac{1}{3x^2+x+1}\\)dx
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\n\u222b\\(\\frac{dx}{\\sqrt{5-2x^2+4x}}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

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

Question 1.
\n\u222b\\(\\frac{x+1}{\\sqrt{x^2-x+1}}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\n\u222b(6x + 5)\\(\\sqrt{6-2x^2+x}\\)dx
\nSolution:
\nLet 6x + 5 = A(1 – 4x) + B
\nEquating the constants
\nA + B = 5
\nB = 5 – A = 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\n\u222b(6x + 5)\\(\\sqrt{6-2x^2+x}\\)dx
\n\"Inter
\n\"Inter<\/p>\n

Question 3.
\n\u222b\\(\\frac{dx}{4+5\\sin x}\\)
\nSolution:
\nt = tan \\(\\frac{x}{2}\\) \u21d2 dt = sec\u00b2 \\(\\frac{x}{2}\\) . \\(\\frac{1}{2}\\)dx
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 4.
\n\u222b\\(\\frac{1}{2-3\\cos 2x}\\)dx
\nSolution:
\nt = tan x \u21d2 dt = sec\u00b2 x dx
\n= (1 + tan\u00b2 x)dx
\n= (1 +t\u00b2)dx
\ndx = \\(\\frac{dt}{1+t^2}\\)
\n\"Inter<\/p>\n

Question 5.
\n\u222bx\\(\\sqrt{1+x-x^2}\\)dx
\nSolution:
\nLet x = A(1 – 2x) + B
\nEquating the coefficients of x
\n1 = -2 A \u21d2 A = –\\(\\frac{1}{2}\\)
\nEquating the constants
\n0 = A + B \u21d2 B = -A = \\(\\frac{1}{2}\\)
\n\u222bx\\(\\sqrt{1+x-x^2}\\)dx
\n= –\\(\\frac{1}{2}\\)\u222b(1 – 2x)\\(\\sqrt{1+x-x^2}\\)dx + \\(\\frac{1}{2}\\)\u222b\\(\\sqrt{1+x-x^2}\\)dx
\n\"Inter<\/p>\n

Question 6.
\n\u222b\\(\\frac{dx}{(1+x)\\sqrt{3+2x-x^2}}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 7.
\n\u222b\\(\\frac{dx}{4\\cos x+3\\sin x}\\)
\nSolution:
\nLet t = tan\\(\\frac{x}{2}\\) so that dx = \\(\\frac{2dt}{1+t^2}\\)
\n\"Inter<\/p>\n

Question 8.
\n\u222b\\(\\frac{1}{\\sin x+\\sqrt{3} \\cos x}\\)dx
\nSolution:
\nLet t = tan \\(\\frac{x}{2}\\) so that dx = \\(\\frac{2dt}{1+t^2}\\)
\nsin x = \\(\\frac{2t}{1+t^2}\\), cos x = \\(\\frac{1-t^2}{1+t^2}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 9.
\n\u222b\\(\\frac{dx}{5+4\\cos 2x}\\).
\nSolution:
\n\"Inter<\/p>\n

Question 10.
\n\u222b\\(\\frac{2\\sin x+3\\cos x+4}{3\\sin x+4\\cos x+5}\\)dx.
\nSolution:
\nLet 2 sin x + 3 cos x + 4
\n= A(3 sin x + 4 cos x + 5) + 3(3 cos x – 4 sin x) + C
\nEquating the co-efficient of
\nsin x, we get 3A – 4B = 2
\ncos x, we get 4A + 3B = 3
\n\"Inter
\nEquating the constants
\n4 = 5A + C
\nC = 4 – 5A = 4 – 5.\\(\\frac{18}{25}\\) = \\(\\frac{2}{5}\\)
\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 11.
\n\u222b\\(\\sqrt{\\frac{5-x}{x-2}}\\) dx on (2, 5).
\nSolution:
\n\"Inter
\nLet 5 – x = A. \\(\\frac{d}{dx}\\)(7x – 10 – x\u00b2) + B
\n\u21d2 5 – x = A(7 – 2x) + B
\nEquating coffs. of like terms
\n-2A = -1 \u21d2 A = \\(\\frac{1}{2}\\)
\n7A + B = 5
\n7(+\\(\\frac{1}{2}\\)) + B = 5 \u21d2 B = 5 – \\(\\frac{7}{2}\\) = \\(\\frac{3}{2}\\)
\n\u2234 5 – x = \\(\\frac{1}{2}\\)(7 – 2x) + \\(\\frac{3}{2}\\)
\n\"Inter
\n\"Inter<\/p>\n

Question 12.
\n\u222b\\(\\sqrt{\\frac{1+x}{1-x}}\\) dx on (-1, 1).
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 13.
\n\u222b\\(\\frac{dx}{(1 – x)\\sqrt{3-2x+x^2}}\\) on (-1, 3).
\nSolution:
\nPut 1 – x = \\(\\frac{1}{1}\\) \u21d2 1 – \\(\\frac{1}{1}\\) = x\\(\\frac{1}{1-x}\\) = t
\ndx = \\(\\frac{1}{t^2}\\)dt
\n3 – 2x – x\u00b2 = 3 – 2(\\(\\frac{1-1}{1}\\)) – (\\(\\frac{1-1}{1}\\))\u00b2
\n\"Inter
\n\"Inter<\/p>\n

Question 14.
\n\u222b\\(\\frac{dx}{(x + 2)\\sqrt{x+1}}\\) on (-1, \u221e).
\nSolution:
\nPut = x + 1 = t\u00b2 \u21d2 dx = 2t dt and
\nx + 2 = 1 + t\u00b2
\n\"Inter<\/p>\n

Question 15.
\n\u222b\\(\\frac{dx}{(2x + 3)\\sqrt{x+2}}\\) on I \u2282 (-2, \u221e)\\{\\(\\frac{-3}{2}\\)}.
\nSolution:
\nPut x + 2 = t\u00b2 \u21d2 dx = 2t dt and
\n2x + 3 = 2(t\u00b2 – 2) + 3 = 2t\u00b2 – 1
\n\"Inter<\/p>\n

Question 16.
\n\u222b\\(\\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}}\\)dx on (0, 1).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 17.
\n\u222b\\(\\frac{dx}{(x + 1)\\sqrt{2x^2+3x+1}}\\) on
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 18.
\n\u222b\\(\\sqrt{e^x-4}\\) dx on [loge<\/sub> 4, \u221e]
\nSolution:
\nPut ex<\/sup> – 4 = t\u00b2 \u21d2 ex<\/sup> dx = 2t dt
\n\"Inter<\/p>\n

Question 19.
\n\u222b\\(\\sqrt{1+\\sec x}\\) dx on [(2n – \\(\\frac{1}{2}\\))\u03c0 – (2n + \\(\\frac{1}{2}\\))\u03c0], (n \u2208 Z).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 20.
\n\u222b\\(\\frac{dx}{1+x^4}\\) on R.
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(d) I. Evaluate the following integrals. Question 1. \u222b Solution: Question 2. \u222bd\u03b8 Solution: Question 3. \u222bdx Solution: t = sin … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25428"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=25428"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25428\/revisions"}],"predecessor-version":[{"id":25478,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25428\/revisions\/25478"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=25428"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=25428"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=25428"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}