{"id":35537,"date":"2022-09-19T17:26:45","date_gmt":"2022-09-19T11:56:45","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35537"},"modified":"2022-09-19T17:26:45","modified_gmt":"2022-09-19T11:56:45","slug":"inter-2nd-year-maths-2b-integration-solutions-ex-6f","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-integration-solutions-ex-6f\/","title":{"rendered":"Inter 2nd Year Maths 2B Integration Solutions Ex 6(f)"},"content":{"rendered":"

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

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

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

Question 1.
\n\u222bex<\/sup>(1 + x\u00b2) dx
\nSolution:
\n\u222bex<\/sup>(1 + x\u00b2) dx = \u222bex<\/sup> + \u222bx\u00b2 . ex<\/sup> dx
\n= ex<\/sup> + (x\u00b2.ex<\/sup> – 2 \u222bx.ex<\/sup>dx)
\n= ex<\/sup> + x\u00b2.ex<\/sup> – 2(x.ex<\/sup> – \u222bex<\/sup> dx
\n= ex<\/sup> + x\u00b2.ex<\/sup> – 2x. ex<\/sup> + 2ex<\/sup> + C
\n= ex<\/sup>(x\u00b2 – 2x + 3) + C<\/p>\n

Question 2.
\n\u222bx\u00b2 e-3x<\/sup> dx
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\n\u222bx\u00b3 eax<\/sup> dx
\nSolution:
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nShow that
\n\u222bxn<\/sup>.e-x<\/sup> dx = -xn<\/sup> e-x<\/sup> + n\u222bxn-1<\/sup>.e-x<\/sup> dx
\nSolution:
\n\u222bxn<\/sup>.e-x<\/sup> dx = \\(\\frac{x^ne^{-x}}{(-1)}\\) + \u222be-x<\/sup>. nxn-1<\/sup> dx
\n= -xn<\/sup> . e-x<\/sup> + n\u222bxn-1<\/sup>e-x<\/sup> dx<\/p>\n

Question 2.
\nIf In<\/sub> = \u222bcosn<\/sup> x dx, than show that
\nIn<\/sub> = \\(\\frac{1}{n}\\) cosn-1<\/sup> x sin x + \\(\\frac{n-1}{n}\\) In-2<\/sub>.
\nSolution:
\nIn<\/sub> = \u222bcosn<\/sup> x dx = \u222bcosn-1<\/sup> x. cos x dx
\n= cosn-1<\/sup>x . sin x – \u222bsin x.(n – 1)cosn-2<\/sup>x (-sin x)dx
\n= cosn-1<\/sup>x . sin x + (n – 1) \u222bcosn-2<\/sup>x(1 – cos\u00b2 x)dx
\n= cosn-1<\/sup>x . sin x + (n – 1)In-2<\/sub> – (n – 1) In<\/sub>
\n\u2234 In<\/sub>(1 + n – 1) = cosn-1<\/sup>x. sin x + (n – 1) In-2<\/sub>
\nIn<\/sub> = \\(\\frac{\\cos ^{n-1} x \\sin x}{n}+\\frac{n-1}{n}\\)In-2<\/sub><\/p>\n

\"Inter<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nObtain reduction formula for In<\/sub> = \u222bcotn<\/sup> x dx, n being a positive integer, n \u2265 2 and deduce the value of \u222bcot4<\/sup> x dx.
\nSolution:
\nIn<\/sub> = \u222bcotn<\/sup> dx = \u222bcotn-2<\/sup> x . cot\u00b2 x dx
\n= \u222bcotn-2<\/sup> x . (cosec\u00b2 x – 1) dx
\n= \u222bcotn-2<\/sup> x . cosec\u00b2 x dx – In-2<\/sub>
\n= – \\(\\frac{cot^{n-1}x}{n-1}\\) – In-2<\/sub>
\nn = 4 \u21d2 I4<\/sub> = –\\(\\frac{cot^{3}x}{3}\\) – I2<\/sub>
\nn = 2 \u21d2 I2<\/sub> = – cot x – I0<\/sub> where I0<\/sub> = \u222bdx = x
\nI2<\/sub> = – cot x – x
\nI4<\/sub> = –\\(\\frac{cot^{3}x}{3}\\) – (-cot x – x) + C
\n= –\\(\\frac{cot^{3}x}{3}\\) + cot x + x + C<\/p>\n

Question 2.
\nObtain the reduction formula for In<\/sub> = \u222bcosecn<\/sup> x dx, n being a positive integer, n \u2265 2 and deduce the value of \u222bcosec5<\/sup> x dx.
\nSolution:
\nIn<\/sub> = \u222bcosecn<\/sup> x dx
\n= cosecn-2<\/sup> x(-cot x) + \u222bcot x. (n – 2)cosecn-3<\/sup> x. (cot x)dx
\n= -cosecn-2<\/sup>x. cot x + (n – 2)\u222bcosecn-2<\/sup>x. (cosec\u00b2 x – 1)dx
\n\"Inter<\/p>\n

Question 3.
\nIf Im, n<\/sub> = \u222bsinm<\/sup> xcosn<\/sup> xdx, then show that
\nfor a positive integer m \u2265 2.
\nSolution:
\nIm, n<\/sub> = \u222b(sinm<\/sup> x) (cosn<\/sup> x) dx
\n= \u222bsinm-1<\/sup> (x).(cos x)n<\/sup>. sin x dx
\n= \u222bsinm-1<\/sup> (x)(cos x)n<\/sup>(-sin x) dx
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

Question 4.
\nEvaluate \u222bsin5<\/sup> x cos4<\/sup> x dx
\nSolution:
\nReduction formula
\nIm, n<\/sub> = \\(\\frac{-\\sin ^{m-1} x \\cdot \\cos ^{n+1} x}{m+n}+\\frac{m-1}{m+n}\\).Im-2, n
\n\"Inter<\/sub><\/p>\n

 <\/p>\n

\"Inter<\/p>\n

Question 5.
\nIf In<\/sub> = \u222b(log x)n<\/sup> dx, then show that In<\/sub> = x(log x)n<\/sup> – nIn-1<\/sub>, and hence find
\n\u222b(log x)4<\/sup> dx.
\nSolution:
\nIn<\/sub> = \u222b(log x)n<\/sup> dx
\n= (log x)n<\/sup>. x – \u222bx . n . (log x)n-1<\/sup> . \\(\\frac{1}{x}\\)dx
\n= x.(log x)n<\/sup> – n\u222b(log x)n-1<\/sup> dx
\n= x(log x)n<\/sup> – n . In-1<\/sub>
\nI4<\/sub> = x(log x)4<\/sup> – 4 . I3<\/sub>
\nI3<\/sub> = x(log x)\u00b3 – 3 . I2<\/sub>
\nI2<\/sub> = x(log x)\u00b2 – 2 . I1<\/sub>
\nI1<\/sub> = x log x I0<\/sub> where I0<\/sub> = \u222bdx = x
\nI1<\/sub> = x log x – x
\nI2<\/sub> = (x (log x)\u00b2 – 2x log x = 2x
\nI3<\/sub> = x(log x)\u00b3 – 3(x (log x)\u00b2 – 2x log x + 2x)
\n= x . (log x)\u00b3 – 3x (log x)\u00b2 + 6x(log x) – 6x
\nI4<\/sub> = x(log x)4<\/sup> – 4[x . (log x)\u00b3 – 3x(log x)\u00b2 + 6x(log x) -6x] + C
\n= x[(log x)4<\/sup> – 4(log x)\u00b3 + 12(log x)\u00b2 – 24(log x) + 24] + C<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(f) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(f) I. Evaluate the following integrals. Question 1. \u222bex(1 + x\u00b2) dx Solution: \u222bex(1 + x\u00b2) dx = \u222bex + \u222bx\u00b2 …<\/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\/35537"}],"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=35537"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35537\/revisions"}],"predecessor-version":[{"id":35538,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35537\/revisions\/35538"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35537"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35537"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35537"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}