{"id":9534,"date":"2024-02-23T09:47:34","date_gmt":"2024-02-23T04:17:34","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9534"},"modified":"2024-02-24T17:54:33","modified_gmt":"2024-02-24T12:24:33","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6c\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(c)"},"content":{"rendered":"

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

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

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

Question 1.
\n\u222bx sec2<\/sup>x dx on I \u2282 R – {\\(\\frac{(2 n+1) \\pi}{2}\\) : n is an integer}.
\nSolution:
\nWe use the formula for integration by parts which state that
\n\"TS<\/p>\n

Question 2.
\n\\(\\int e^x\\left(\\tan ^{-1} x+\\frac{1}{1+x^2}\\right) d x\\), x \u2208 R.
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\n\\(\\int \\frac{\\log x}{x^2} d x\\) on(0, \u221e).
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 4.
\n\u222b(log x)2<\/sup> dx on (0, \u221e).
\nSolution:
\nTake (log x)2<\/sup> = u and 1 = v.
\nThen by integration by parts,
\n\u222b(log x)2<\/sup> . 1 . dx = (log x)2<\/sup> . x – \u222b2 (log x) \\(\\frac{1}{x}\\) . x dx
\n= x(log x)2<\/sup> – 2 \u222blog x . 1 . dx
\n= x(log x)2<\/sup> – 2[logx . x – \u222b\\(\\frac{1}{x}\\) . x dx]
\n= x(log x)2<\/sup> – 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:
\nLet f(x) = sec x then f'(x) = sec x tan x
\n\u2234 Using \u222bex<\/sup> [f(x) + f'(x)] dx = ex<\/sup> f(x) + c
\nwe have \u222bex<\/sup> (sec x + sec x tan x) dx = ex<\/sup> sec x + c<\/p>\n

Question 6.
\n\u222bex<\/sup> cos x dx on R.
\nSolution:
\nLet I = \u222bex<\/sup> cos x dx
\nand take u = ex<\/sup> and v = cos x.
\nThen using integration by parts,
\nI = ex<\/sup> (sin x) – \u222bex<\/sup> sin x dx
\n= ex<\/sup> (sin x) – [ex<\/sup>(-cos x) – \u222bex<\/sup> (-cos x) dx]
\n= ex<\/sup> sin x + ex<\/sup> cos x – \u222bex<\/sup> cos x dx
\n= ex<\/sup> (sin x + cos x) – 1
\n2I = ex<\/sup> (sin x + cos x)
\nI = \\(\\frac{1}{2}\\) ex<\/sup> (sin x + cos x) + c
\n\u2234 \u222bex<\/sup> cos x dx = \\(\\frac{1}{2}\\) ex<\/sup> (sin x + cos x)<\/p>\n

Question 7.
\n\u222bex<\/sup> (sin x + cos x) dx on R.
\nSolution:
\nTake f(x) = sin x then f'(x) = cos x
\nSo by using formula \u222bex<\/sup> [f(x) + f'(x)] dx = ex<\/sup> f(x) + c
\nwe have \u222bex<\/sup> (sin x + cos x) dx = ex<\/sup> sin x + c.<\/p>\n

\"TS<\/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:
\nTake f(x) = log|sec x| then f'(x) = \\(\\frac{1}{\\sec x}\\) (sec x tan x) = tan x
\nSo by the formula \u222bex<\/sup> [f(x) + f'(x)] dx = ex<\/sup> f(x) + c
\nwe have \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 number and n \u2260 -1.
\nSolution:
\nTake u = log x and v = xn<\/sup>
\napplying integration by parts,
\n\"TS
\n\"TS<\/p>\n

Question 2.
\n\u222blog(1 + x2<\/sup>) dx on R.
\nSolution:
\nTake log(1 + x2<\/sup>) = u and 1 = v then
\nusing integration by parts, we get
\n\"TS<\/p>\n

Question 3.
\n\u222b\u221ax log x dx on (0, \u221e).
\nSolution:
\nTake u = log x and v = x1\/2<\/sup> and
\nusing integration by parts, we get
\n\"TS<\/p>\n

Question 4.
\n\\(\\int e^{\\sqrt{x}} d x\\) on (0, \u221e).
\nSolution:
\n\"TS<\/p>\n

Question 5.
\n\u222bx2<\/sup> cos x dx on R.
\nSolution:
\nTake x2<\/sup> = u and cos x = v,
\nand using integration by parts, we get
\n\u222bx2<\/sup> cos x dx = x2<\/sup> (sin x) – \u222b2x sin x dx
\n= x2<\/sup> sin x – [2x(-cos x) – \u222b2 . (-cos x) dx]
\n= x2<\/sup> sin x + 2x cos x – 2\u222bcos x dx
\n= x2<\/sup> sin x + 2x cos x – 2 sin x + c<\/p>\n

\"TS<\/p>\n

Question 6.
\n\u222bx sin2<\/sup>x dx on R.
\nSolution:
\n\"TS<\/p>\n

Question 7.
\n\u222bx cos2<\/sup>x dx on R.
\nSolution:
\n\"TS<\/p>\n

Question 8.
\n\u222bcos\u221ax dx on R.
\nSolution:
\n\u222bcos\u221ax dx = \u222b\\(\\frac{1}{\\sqrt{x}}\\) \u221ax cos\u221ax dx
\nTaking \u221ax = t, we get \\(\\frac{1}{2 \\sqrt{x}}\\) dx = dt
\n\u21d2 \\(\\frac{\\mathrm{dx}}{\\sqrt{\\mathrm{x}}}\\) = 2 dt
\n\u2234 \u222bcos \u221ax dx = 2\u222bt cos t dt
\nusing Integration by parts,
\n= 2[t(sin t) – \u222b1 . sin t dt]
\n= 2[t sin t + cos t + c]
\n= 2[\u221ax sin \u221ax + cos \u221ax ] + c<\/p>\n

Question 9.
\n\u222bx sec2<\/sup>2x dx on I \u2282 R \\ {(2n\u03c0 + 1) \\(\\frac{\\pi}{4}\\) : n \u2208 Z}
\nSolution:
\nTaking x = u and sec2<\/sup>2x = v,
\nand applying integration by parts we get
\n\u222bx sec2<\/sup>2x dx = x(\\(\\frac{1}{2}\\) tan 2x – \u222b1 . \\(\\frac{1}{2}\\) tan 2x dx
\n= \\(\\frac{x}{2}\\) tan 2x – \\(\\frac{1}{2}\\) \u222btan 2x dx
\n= \\(\\frac{x}{2}\\) tan 2x – \\(\\frac{1}{4}\\) log|sec 2x| + c<\/p>\n

Question 10.
\n\u222bx cot2<\/sup>x dx on I \u2282 R \\ {n\u03c0 : n \u2208 Z).
\nSolution:
\n\u222bx cot2<\/sup>x dx = \u222bx(cosec2<\/sup>x – 1) dx = \u222bx cosec2<\/sup>x dx – \u222bx dx
\nTaking u = x and v = cosec2<\/sup>x on the first integral
\nand using integration by parts we get
\n\u222bx cot2<\/sup>x dx = x(-cot x) – \u222b1 . (-cot x) dx – \\(\\frac{x^2}{2}\\)
\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 + sec2<\/sup>x) dx on I \u2282 R \\ {(2n + 1)\\(\\frac{\\pi}{2}\\) : n \u2208 Z}.
\nSolution:
\nLet f(x) = tan x, then f'(x) = sec2<\/sup>x
\nSo by the formula \u222bex<\/sup> [f(x) + f'(x)] dx = ex<\/sup> f(x) + c
\nwe have \u222bex<\/sup> (tan x + sec2<\/sup>x) dx = ex<\/sup> tan x + c<\/p>\n

Question 12.
\n\u222b\\(e^x\\left(\\frac{1+x \\log x}{x}\\right) d x\\) on (0, \u221e). (Mar. ’13)
\nSolution:
\n\"TS<\/p>\n

Question 13.
\n\u222beax<\/sup>\u00a0sin bx dx on R, a, b \u2208 R.
\nSolution:
\nLet I = \u222beax<\/sup> sin bx dx
\nTaking u = eax<\/sup> and v = sin bx
\nand applying integration by parts
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

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

Question 15.
\n\\(\\int \\frac{d x}{\\left(x^2+a^2\\right)^2}\\), (a > 0) on R.
\nSolution:
\nTake substitution x = a tan \u03b8
\nso that dx = a sec2<\/sup>\u03b8 d\u03b8
\n\"TS<\/p>\n

Question 16.
\n\u222bex<\/sup> log(e2x<\/sup> + 5ex<\/sup> + 6) dx on R.
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 17.
\n\\(\\int e^x \\frac{(x+2)}{(x+3)^2} d x\\) on I \u2282 R \\ {-3}.
\nSolution:
\n\"TS<\/p>\n

Question 18.
\n\u222bcos(log x) dx on (0, \u221e).
\nSolution:
\nLet I = \u222bcos (log x) dx = \u222bcos (log x) . 1 dx
\nTake u = cos (log x) and v = 1
\nand using integration by parts successively.
\nI = cos (log x) . x – \u222b-sin (log x) \\(\\frac{1}{x}\\) . x . dx
\n= x cos (log x) + \u222bsin(log x) dx
\n= x cos (log x) + sin (log x) . x – \u222bcos (log x) \\(\\frac{1}{x}\\) . x . dx
\n= x cos (log x) + x . sin(log x) – \u222bcos (log x) dx
\n= x [cos (log x) + sin (log x)] – \u222bcos (log x) dx
\n= x [cos (log x) + sin (log x)] – I
\n\u2234 2I = x [cos (log x) + sin (log x)]
\n\u21d2 I = \\(\\frac{x}{2}\\) [cos (log x) + sin (log x)] + c
\n\u2234 \u222bcos (log x) dx = \\(\\frac{x}{2}\\) [cos (log x) + sin (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:
\nLet u = tan-1<\/sup>x and v = x then using integration by parts
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\n\u222bx2<\/sup> tan-1<\/sup>x dx, x \u2208 R.
\nSolution:
\nTake u = tan-1<\/sup>x and v = x2<\/sup>
\nand apply integration by parts we get
\n\"TS
\n\"TS<\/p>\n

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

Question 4.
\n\u222bx cos-1<\/sup>x dx, x \u2208 (-1, 1).
\nSolution:
\nLet cos-1<\/sup>x = \u03b8 then cos \u03b8 = x
\n\u21d2 dx = -sin \u03b8 d\u03b8
\n\u2234 \u222bx cos-1<\/sup>x dx = \u222b\u03b8 cos \u03b8 (-sin \u03b8 d\u03b8) – \\(\\frac{1}{2}\\) \u222b\u03b8 sin 2\u03b8 d\u03b8
\nUsing integration by parts by taking u = \u03b8, v = sin 2\u03b8 we get
\n\"TS<\/p>\n

Question 5.
\n\u222bx2<\/sup> sin-1<\/sup>x dx, x \u2208 (-1, 1).
\nSolution:
\nLet sin-1<\/sup>x = \u03b8 then sin \u03b8 = x
\n\u21d2 dx = cos \u03b8 d\u03b8
\n\u2234 \u222bx2<\/sup> sin-1<\/sup>x dx = \u222b\u03b8 sin2<\/sup>\u03b8 cos \u03b8 d\u03b8
\nUsing integration by parts
\nby choosing functions u = \u03b8 and v = sin2<\/sup>\u03b8 cos \u03b8, we get
\n\"TS<\/p>\n

Question 6.
\n\u222bx log(1 + x) dx, x \u2208 (-1, \u221e).
\nSolution:
\nTake u = log(1 + x) and v = x
\nand apply integration by parts
\n\"TS
\n\"TS<\/p>\n

Question 7.
\n\u222bsin\u221ax dx on (0, \u221e).
\nSolution:
\n\\(\\int \\frac{\\sqrt{x}}{\\sqrt{x}} \\sin \\sqrt{x} d x\\)
\nLet \u221ax = t then \\(\\frac{1}{2 \\sqrt{x}}\\) dx = dt
\n\u21d2 \\(\\frac{d x}{\\sqrt{x}}\\) = 2 dt
\n\u2234 \u222bsin \u221ax dx = 2\u222bt sin t dt,
\nusing Integration by parts by taking u = t and v = sin t, we get
\n= 2[t(-cos t) – \u222b1 . (-cos t) dt]
\n= 2[-t cos t + \u222bcos t dt]
\n= 2[-t cos t + sin t] + c
\n= 2[sin \u221ax – \u221ax cos \u221ax ] + c<\/p>\n

\"TS<\/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, taking u = eax<\/sup> and v = sin (bx + c)
\nand applying integration by parts,
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 9.
\n\u222bax<\/sup> cos 2x dx on R (a > 0 and a \u2260 1).
\nSolution:
\nLet I = \u222bax<\/sup> cos 2x dx
\nusing integration by parts by taking cos 2x = u and ax<\/sup> = v, we have
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 10.
\n\\(\\int \\tan ^{-1}\\left(\\frac{3 x-x^3}{1-3 x^2}\\right) d x\\) on I \u2282 R – \\(\\left\\{-\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}\\right\\}\\).
\nSolution:
\n\"TS<\/p>\n

Question 11.
\n\u222bsinh-1<\/sup>x dx on R.
\nSolution:
\nTake sinh-1<\/sup>x = u and v = 1,
\nApplying integration by parts we get
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 12.
\n\u222bcosh-1<\/sup>x dx on [1, \u221e).
\nSolution:
\n\u222bcosh-1<\/sup>x dx = \u222bcosh-1<\/sup>x . 1 . dx
\nTake u = cosh-1<\/sup>x and v = 1 then
\n\"TS<\/p>\n

Question 13.
\n\u222btanh-1<\/sup>x dx on (-1, 1).
\nSolution:
\n\u222btanh-1<\/sup>x dx = \u222btanh-1<\/sup>x . 1 . dx
\nTake u = tanh-1<\/sup>x and v = 1
\nand apply integration by parts we get
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 6 Integration Ex 6(c) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(c) I. Evaluate the following integrals. Question 1. \u222bx sec2x dx on I \u2282 R – { : n is an … 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\/9534"}],"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=9534"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9534\/revisions"}],"predecessor-version":[{"id":9581,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9534\/revisions\/9581"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9534"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9534"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9534"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}