{"id":35531,"date":"2022-09-19T17:24:19","date_gmt":"2022-09-19T11:54:19","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35531"},"modified":"2022-09-19T17:24:31","modified_gmt":"2022-09-19T11:54:31","slug":"inter-2nd-year-maths-2b-integration-solutions-ex-6c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-integration-solutions-ex-6c\/","title":{"rendered":"Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)"},"content":{"rendered":"

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

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

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

Question 1.
\n\u222bx sec\u00b2 x dx on I \u2282 R\\{\\(\\frac{(2n+1)\\pi}{2}\\) : n is integer}.
\nSolution:
\n\u222buvdx = u\u222bvdx – \u222b[\\(\\frac{d}{du}\\)(U).\u222bvdx]dx
\nLet v = sec\u00b2 x and u = x then
\n\u222bx sec\u00b2 x dx = x(tan x) – \u222btan x dx
\n= x tan x – log|sec x| + C<\/p>\n

Question 2.
\n\u222bex<\/sup>(tan-1<\/sup> x + \\(\\frac{1}{1+x^2}\\))dx, x \u2208 R.
\nSolution:
\n\u222bex<\/sup>[f(x) + f'(x)] dx = ex<\/sup>. f(x) + C
\nLet f(x) = tan-1<\/sup> x so that f'(x) = \\(\\frac{1}{1+x^2}\\)
\n\u2234 \u222bex<\/sup>(tan-1<\/sup> x + \\(\\frac{1}{1+x^2}\\))dx = ex<\/sup>tan-1<\/sup> x + C<\/p>\n

Question 3.
\n\u222b\\(\\frac{log x}{x^2}\\) dx on (0, \u221e).
\nSolution:
\n\u222b\\(\\frac{log x}{x^2}\\) dx = (log x)(-\\(\\frac{1}{x}\\)) + \u222b\\(\\frac{1}{x}\\) . \\(\\frac{1}{x}\\)dx
\n= – \\(\\frac{1}{x}\\)log x – \\(\\frac{1}{x}\\) + C<\/p>\n

Question 4.
\n\u222b(log x)\u00b2 dx (0, \u221e).
\nSolution:
\n\u222b(log x)\u00b2 dx = (log x)\u00b2 x – \u222bx . 2log x.\\(\\frac{1}{x}\\) dx
\n= x (log x)\u00b2 – 2 \u222blog x dx
\n= x (log x)\u00b2 – 2(x . log x – \u222bx\\(\\frac{1}{x}\\)dx)
\n= x(log x)\u00b2 – 2x. log x + 2x + c<\/p>\n

Question 5.
\n\u222bex<\/sup>(sec x + sec x . tan x)dx on I \u2282 R\\{(2n + 1)\\(\\frac{\\pi}{2}\\) : n \u2208 Z}.
\nSolution:
\n\u222bex<\/sup>(sec x + sec x . tan x)dx = ex<\/sup>.sec x + C
\n[\u222bex<\/sup>[f(x) + f'(x)]dx = ex<\/sup> f(x) + C]<\/p>\n

\"Inter<\/p>\n

Question 6.
\n\u222bex<\/sup> cos x dx on R.
\nSolution:
\nI = \u222bex<\/sup> cos x dx = ex<\/sup> sin x – \u222bsin x . ex<\/sup> dx
\n= ex<\/sup>. sin x + ex<\/sup>. cos x – \u222bex<\/sup> . cos x dx
\n= ex<\/sup>(sin x + cos x) – I
\n2I = ex<\/sup>(sin x + cos x)
\nI = \\(\\frac{e^x}{2}\\)(sin x + cos x) + C<\/p>\n

Question 7.
\n\u222bex<\/sup>(sin x + cos x) dx on R.
\nSolution:
\n\u222bex<\/sup>(sin x + cos x) dx
\nf(x) = sin x \u21d2 f'(x) = cos x
\n\u2234 \u222bex<\/sup>(sin x + cos x) dx = ex<\/sup>. sin x + C<\/p>\n

Question 8.
\n\u222b(tan x + log sec x)ex<\/sup> dx on ((2n – \\(\\frac{1}{2}\\))\u03c0, (2n + \\(\\frac{1}{2}\\))\u03c0) n \u2208 Z.
\nSolution:
\nt = log|sec| \u21d2 dt = \\(\\frac{1}{\\sec x}\\). sec x .tan x dx
\n= tan x dx
\n\u222b(tan x + log sec x)ex<\/sup> dx = ex<\/sup>.log|sec x| + C<\/p>\n

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

Question 1.
\n\u222bxn<\/sup>log x dx on (0, \u221e), n is a real nember and n \u2260 -1.
\nSolution:
\n\u222bxn<\/sup>log x dx
\n\"Inter<\/p>\n

Question 2.
\n\u222blog (1 + x\u00b2) dx on R.
\nSolution:
\n\u222blog (1 + x\u00b2) dx on
\n= [log (1 + x\u00b2). x – \u222bx\\(\\frac{1}{1+x^2}\\)2x dx
\n= x log (1 + x\u00b2) – 2\u222b\\(\\frac{1+x^2-1}{1+x^2}\\)dx
\n= x log (1 + x\u00b2) – 2\u222bdx + 2\u222b\\(\\frac{dx}{1+x^2}\\)
\n= x log (1 + x\u00b2) – 2x + 2 tan-1<\/sup> x + C<\/p>\n

Question 3.
\n\u222b\\(\\sqrt{x}\\) log x dx on (0, \u221e).
\nSolution:
\n\u222b\\(\\sqrt{x}\\) log x dx
\n\"Inter<\/p>\n

Question 4.
\n\u222be\u221ax<\/sup> dx on (0, \u221e).
\nSolution:
\nt = \u221ax \u21d2 x = t\u00b2
\ndx = 2t dt
\n\u222be\u221ax<\/sup> dx = 2\u222bt et<\/sup> dt
\n= 2 [t et<\/sup> – \u222bet<\/sup> dt
\n= 2 (t et<\/sup> – et<\/sup>) + C
\n= 2\u221ax e\u221ax<\/sup> – 2e\u221ax<\/sup> + C<\/p>\n

Question 5.
\n\u222bx\u00b2 cos x dx on R.
\nSolution:
\n\u222bx\u00b2 cosx dx = x\u00b2(sin x) – \u222bsin x(2x dx)
\n= x\u00b2 sin x + 2\u222bx(-sin x)dx
\n= x\u00b2. sin x + 2[x cos x – \u222bcos x dx]
\n= x\u00b2 sin x + 2x cos x – 2 sin + C<\/p>\n

\"Inter<\/p>\n

Question 6.
\n\u222bx sin\u00b2 x dx on R.
\nSolution:
\n\u222bx sin\u00b2 x dx = \\(\\frac{1}{2}\\)\u222bx(1 – cos 2x) dx
\n= \\(\\frac{1}{2}\\)[\u222bx dx – \u222bx cos 2x dx]
\n= \\(\\frac{1}{2}\\)[\\(\\frac{x^2}{2}\\) – {x . \\(\\frac{\\sin 2x}{2}\\) – \\(\\frac{1}{2}\\)\u222bsin 2x dx}]
\n= \\(\\frac{x^2}{4}\\) – \\(\\frac{x}{4}\\)sin 2x + \\(\\frac{1}{4}\\)\u222bsin 2x dx
\n= \\(\\frac{x^2}{4}\\) – \\(\\frac{x}{4}\\)sin 2x – \\(\\frac{1}{8}\\)cos 2x + C<\/p>\n

Question 7.
\n\u222bx cos\u00b2 x dx on R.
\nSolution:
\n\u222bx cos\u00b2 x dx = \\(\\frac{1}{2}\\)\u222bx(1+ cos 2x)dx
\n= \\(\\frac{1}{2}\\)[\u222bx dx + \u222bx cos 2x dx]
\n= \\(\\frac{1}{2}\\)[\\(\\frac{x^2}{2}\\) + {x.\\(\\frac{\\sin 2x}{2}\\) – \\(\\frac{1}{2}\\)\u222bsin 2x dx}]
\n= \\(\\frac{x^2}{4}\\) + \\(\\frac{x}{4}\\) sin 2x – \\(\\frac{1}{4}\\) \u222bsin 2x dx
\n= \\(\\frac{x^2}{4}\\) + \\(\\frac{x}{4}\\) sin 2x + \\(\\frac{1}{8}\\)cos 2x + C<\/p>\n

Question 8.
\n\u222bcos \u221ax dx on R.
\nSolution:
\nt = x = t\u00b2 \u21d2 dx = 2t dt
\nI = 2\u222bt. cos t dt = 2 (t sin t – \u222bsin t dt)
\n= 2(t sin t + cos t) + C
\n= 2\u221ax sin \u221ax + 2 cos\u221ax + C<\/p>\n

Question 9.
\n\u222bx sec\u00b2 2x dx on I \u2282 R\\ {(2n\u03c0 + 1)\\(\\frac{\\pi}{4}\\) : n \u2208 Z}.
\nSolution:
\n\u222bx sec\u00b2 2x dx = x\\(\\frac{\\tan 2x}{2}\\) – \\(\\frac{1}{2}\\)\u222btan 2x dx
\n= x\\(\\frac{\\tan 2x}{2}\\) – \\(\\frac{1}{2}\\) . \\(\\frac{1}{2}\\) log|sec 2x| + C
\n= x\\(\\frac{\\tan 2x}{2}\\) – \\(\\frac{1}{4}\\) log|sec 2x| + C<\/p>\n

Question 10.
\n\u222bx cot\u00b2 x dx on I \u2282 R\\{n\u03c0 : n \u2208 Z}.
\nSolution:
\n\u222bx cot\u00b2 x dx
\n= \u222bx (cosec\u00b2 x – 1)dx
\n= \u222bx cosec\u00b2x dx – \u222bx dx
\n= x (- cot x) + \u222bcot x dx – \\(\\frac{x^2}{2}\\)
\n= – x cot x + log|sin x| – \\(\\frac{x^2}{2}\\) + C<\/p>\n

Question 11.
\n\u222bex<\/sup>(tan x + sec\u00b2 x)dx on I \u2282 R\\{(2n + 1)\\(\\frac{\\pi}{2}\\) : n \u2208 Z}.
\nSolution:
\nf(x) = tan x \u21d2 f'(x) = sec\u00b2 x dx
\nI = \u222bex<\/sup>[f(x) + f'(x)] dx = ex<\/sup>. f(x) + C
\n= ex<\/sup>. tan x + C<\/p>\n

Question 12.
\n\u222bex<\/sup>(\\(\\frac{1+log x}{x}\\))dx on (0, \u221e).
\nSolution:
\n\u222bex<\/sup>(\\(\\frac{1+log x}{x}\\))dx = \u222bex<\/sup>(log x + \\(\\frac{1}{x}\\))dx
\n= ex<\/sup>.log x + C<\/p>\n

Question 13.
\n\u222beax<\/sup>sin bx dx on R, a, b \u2208 R.
\nSolution:
\nLet I = \u222beax<\/sup>sin bx dx …………. (1)
\n\"Inter<\/p>\n

Question 14.
\n\u222b\\(\\frac{x.e^2}{(x+1)^2}\\) dx on I \u2282 R\\{-1}
\nSolution:
\n\"Inter<\/p>\n

Question 15.
\n\u222b\\(\\frac{dx}{(x^2+a^2)^2}\\), (a > 0) on R.
\nSolution:
\nPut x = tan t
\nThen dx = a sec\u00b2 t dt
\n\"Inter<\/p>\n

Question 16.
\n\u222bex<\/sup> log (e2x<\/sup> + 5ex<\/sup> + 6) dx on R.
\nSolution:
\n\u222bex<\/sup> log (e2x<\/sup> + 5ex<\/sup> + 6) dx
\n\u2235 e2x<\/sup> + 5ex<\/sup> + 6 = (ex<\/sup> + 2)(ex<\/sup> + 3)
\n= \u222bex<\/sup>. log((ex<\/sup> + 2)(ex<\/sup> + 3))dx
\n= \u222bex<\/sup>.{log(ex<\/sup> + 2) + log(ex<\/sup> + 3)}dx
\n= \u222bex<\/sup>log (ex<\/sup> + 2)dx + \u222bex<\/sup>log (ex<\/sup> + 3)dx
\nPut ex<\/sup> = t \u21d2 ex<\/sup>dx = dt
\n= \u222blog(t + 2)dt + \u222blog(t + 3) dt
\n= log(t + 2)\u222b1dt – \u222b{\\(\\frac{d}{dt}\\)log(t + 2). \u222b1dt}dt + log(t + 3)\u222b1dt – \u222b{\\(\\frac{d}{dt}\\)log(t + 3). \u222b1dt}dt
\n= t log(t + 2) – \u222b\\(\\frac{1}{t + 2}\\))t dt + t log(t+ 3) – \u222b\\(\\frac{1}{t + 3}\\)). t dt
\n= t {log(t + 2) + log (t + 3)} – \u222b\\(\\frac{t}{t + 2}\\)dt – \u222b\\(\\frac{t}{t + 3}\\)dt
\n= t log (t\u00b2 + 5t + 6) – \u222b(\\(\\frac{t+2-2}{t + 2}\\))dt – \u222b(\\(\\frac{t+3-3}{t + 3}\\))dt
\n= t. log (t\u00b2 + 5t + 6) – \u222b{1 – \\(\\frac{2}{t + 2}\\)}dt – \u222b{1 – \\(\\frac{3}{t + 3}\\)}dt
\n= t log (t\u00b2 + 5t + 6) – t + 2 log|t + 2| – t + 3 log|t + 3| + C
\n= t log (t\u00b2 + 5t + 6) – 2t + 2 log|t + 2| + 3log|t + 3| + C
\n= ex<\/sup> log(e2x<\/sup> + 5ex<\/sup> + 6) – 2ex<\/sup> + 2log ex<\/sup> + 2 + 3 log ex<\/sup> + 3 + C<\/p>\n

\"Inter<\/p>\n

Question 17.
\n\u222bex<\/sup>.\\(\\frac{x+2}{(x+3)^2}\\) dx on I \u2282 R\\ {-3}
\nSolution:
\n\"Inter<\/p>\n

Question 18.
\n\u222bcos (log x) dx on (0, \u221e).
\nSolution:
\nPut log x = t
\nx = et<\/sup>
\ndx = et<\/sup> . dt
\nI = \u222bet<\/sup> . cos t . dt
\n= et<\/sup> sin t – \u222bsin.et<\/sup> dt
\n= et<\/sup> . sin t + cos t . et<\/sup> – \u222bet<\/sup> . cos t dt
\n2I = et<\/sup> . (sin t + cos t)
\nI = \\(\\frac{e^t}{2}\\)(sin t + cos t)
\n= \\(\\frac{x}{2}\\)[sin(log x) + cos(log x)] + C<\/p>\n

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

Question 1.
\n\u222bx tan-1<\/sup> x dx, x \u2208 R
\nSolution:
\n\u222bx tan-1<\/sup> x dx = (tan-1<\/sup> x)\\(\\frac{x^2}{2}\\) – \\(\\frac{1}{2}\\)\u222bx\u00b2.\\(\\frac{1}{1+x^2}\\)dx
\n\"Inter<\/p>\n

Question 2.
\n\u222bx\u00b2 tan-1<\/sup> x dx, x \u2208 R.
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n\u222b\\(\\frac{tan^{-1}x}{x^2}\\) dx, x \u2208 I \u2282 R\\{0}
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\n\u222bx cos-1<\/sup>x dx, x \u2208 (-1, 1).
\nSolution:
\n\u222bx cos-1<\/sup>x dx
\n\"Inter
\n\"Inter<\/p>\n

Question 5.
\n\u222bx\u00b2 sin-1<\/sup>x dx, x \u2208 (-1, 1).
\nSolution:
\n\u222bx\u00b2 sin-1<\/sup>x dx
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 6.
\n\u222bx log(1 + x) dx, x \u2208 (-1, \u221e).
\nSolution:
\n\u222bx log(1 + x) dx
\n\"Inter
\n\"Inter<\/p>\n

Question 7.
\n\u222bsin \u221ax dx, on (0, \u221e).
\nSolution:
\nx = t\u00b2 \u21d2 dx = 2t dt
\n\u222bsin \u221ax dx = 2\u222bt. sin t dt
\n= 2(t(-cos t) + \u222bcos t dt)
\n= -2t cos t + 2 sin t
\n= -2\u221ax cos \u221ax + 2 sin \u221ax + C<\/p>\n

Question 8.
\n\u222beax<\/sup> sin (bx + c)dx, (a, b, c \u2208 R, b \u2260 0) on R.
\nSolution:
\nLet I = \u222beax<\/sup> sin (bx + c)dx
\n\"Inter
\n\"Inter<\/p>\n

Question 9.
\n\u222bax<\/sup> cos 2x dx on R (a > 0 and a \u2260 0).
\nSolution:
\n\"Inter<\/p>\n

Question 10.
\n\u222btan-1<\/sup>(\\(\\frac{3x-x^3}{1-3x^2}\\)) dx on I \u2282 R\\{-\\(\\frac{1}{\\sqrt{3}}\\), \\(\\frac{1}{\\sqrt{3}}\\)}.
\nSolution:
\nPut x = tan t \u21d2 dx = sec\u00b2 t dt
\nThen
\n\u222btan-1<\/sup>(\\(\\frac{3x-x^3}{1-3x^2}\\)) dx
\n= \u222btan-1<\/sup>(\\(\\frac{3 \\tan t -\\tan^3 t}{1-3\\tan^2 t}\\)) dx
\n= \u222btan-1<\/sup>(tan 3t).sec\u00b2 t dt
\n= 3\u222bt sec\u00b2 t dt
\n= 3[t\u222bsec\u00b2 t dt – \u222b{\\(\\frac{d}{dt}\\)(t) \u222bsec\u00b2 t dt}dt]
\n= 3[t(tan t) – \u222b(1) tan t dt]
\n= 3(x tan-1<\/sup> x – log\\(\\sqrt{1+x^2}\\)) + C
\n= 3x[tan-1<\/sup> x – \\(\\frac{3}{2}\\)log (1 + x\u00b2) + C
\n= 3x tan-1<\/sup>(x) – \\(\\frac{3}{2}\\)log (1 + x\u00b2) + C<\/p>\n

Question 11.
\n\u222bsinh-1<\/sup> x dx on R.
\nSolution:
\n\"Inter<\/p>\n

Question 12.
\n\u222bcosh-1<\/sup> x dx on (1, \u221e).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 13.
\n\u222btanh-1<\/sup>x dx on (-1, 1).
\nSolution:
\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(c) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(c) I. Evaluate the following integrals. Question 1. \u222bx sec\u00b2 x dx on I \u2282 R\\{ : n is integer}. Solution: …<\/p>\n","protected":false},"author":5,"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.guru\/wp-json\/wp\/v2\/posts\/35531"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35531"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35531\/revisions"}],"predecessor-version":[{"id":35532,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35531\/revisions\/35532"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35531"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35531"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35531"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}