{"id":6812,"date":"2021-10-21T17:56:07","date_gmt":"2021-10-21T12:26:07","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6812"},"modified":"2021-11-01T19:32:40","modified_gmt":"2021-11-01T14:02:40","slug":"prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/","title":{"rendered":"Prove that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2."},"content":{"rendered":"

Solution :<\/h2>\n

Let I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx\u00a0 \u00a0 …….(i)<\/p>\n

then I = \\(\\int_{0}^{\\pi\/2}\\) \\(log(sin({\\pi\\over 2}-x))\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx\u00a0 \u00a0 \u00a0…….(ii)<\/p>\n

Adding (i) and (ii), we get<\/p>\n

2I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx + \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = \\(\\int_{0}^{\\pi\/2}\\) (log(sinx)dx + log(cosx))<\/p>\n

\\(\\implies\\) \\(\\int_{0}^{\\pi\/2}\\) log(sinxcosx)dx = \\(\\int_{0}^{\\pi\/2}\\) \\(log({2sinxcosx\\over 2})\\)dx<\/p>\n

= \\(\\int_{0}^{\\pi\/2}\\) \\(log({sin2x\\over 2})\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – \\(\\int_{0}^{\\pi\/2}\\) log(2)dx<\/p>\n

= \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – (log 2)\\({(x)}^{\\pi\/2}_{0}\\)<\/p>\n

\\(\\implies\\)\u00a0 2I = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – \\(\\pi\\over 2\\)log 2\u00a0 \u00a0…(iii)<\/p>\n

Let \\(I_1\\) = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx,\u00a0 putting 2x = t, we get<\/p>\n

\\(I_1\\) = \\(\\int_{0}^{\\pi}\\) log(sint)\\(dt\\over 2\\) = \\(1\\over 2\\) \\(\\int_{0}^{\\pi}\\) log(sint)dt = \\(1\\over 2\\) 2\\(\\int_{0}^{\\pi\/2}\\) log(sint)dt<\/p>\n

\\(I_1\\) = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx<\/p>\n

\\(\\therefore\\)\u00a0 (iii) becomes; 2I = I – \\(\\pi\\over 2\\)log 2<\/p>\n

Hence\u00a0 \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = – \\(\\pi\\over 2\\)log 2<\/p>\n

\n

Similar Questions<\/h3>\n

What is the integration of x tan inverse x dx ?<\/a><\/p>\n

What is the integration of tan inverse root x ?<\/a><\/p>\n

Evaluate : \\(\\int\\) \\(cos^4xdx\\over {sin^3x{(sin^5x + cos^5x)^{3\\over 5}}}\\)<\/a><\/p>\n

Evaluate : \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\)<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx\u00a0 \u00a0 …….(i) then I = \\(\\int_{0}^{\\pi\/2}\\) \\(log(sin({\\pi\\over 2}-x))\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx\u00a0 \u00a0 \u00a0…….(ii) Adding (i) and (ii), we get 2I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx + \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = \\(\\int_{0}^{\\pi\/2}\\) (log(sinx)dx + log(cosx)) \\(\\implies\\) \\(\\int_{0}^{\\pi\/2}\\) log(sinxcosx)dx = \\(\\int_{0}^{\\pi\/2}\\) \\(log({2sinxcosx\\over 2})\\)dx = \\(\\int_{0}^{\\pi\/2}\\) \\(log({sin2x\\over 2})\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – \\(\\int_{0}^{\\pi\/2}\\) log(2)dx …<\/p>\n

Prove that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[52,43],"tags":[],"yoast_head":"\nProve that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2.<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2.\" \/>\n<meta property=\"og:description\" content=\"Solution : Let I = (int_{0}^{pi\/2}) log(sinx)dx\u00a0 \u00a0 …….(i) then I = (int_{0}^{pi\/2}) (log(sin({piover 2}-x)))dx = (int_{0}^{pi\/2}) log(cosx)dx\u00a0 \u00a0 \u00a0…….(ii) Adding (i) and (ii), we get 2I = (int_{0}^{pi\/2}) log(sinx)dx + (int_{0}^{pi\/2}) log(cosx)dx = (int_{0}^{pi\/2}) (log(sinx)dx + log(cosx)) (implies) (int_{0}^{pi\/2}) log(sinxcosx)dx = (int_{0}^{pi\/2}) (log({2sinxcosxover 2}))dx = (int_{0}^{pi\/2}) (log({sin2xover 2}))dx = (int_{0}^{pi\/2}) log(sin2x)dx – (int_{0}^{pi\/2}) log(2)dx … Prove that (int_{0}^{pi\/2}) log(sinx)dx = (int_{0}^{pi\/2}) log(cosx)dx = -(piover 2)log 2. Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\" \/>\n<meta property=\"og:site_name\" content=\"Mathemerize\" \/>\n<meta property=\"article:published_time\" content=\"2021-10-21T12:26:07+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-11-01T14:02:40+00:00\" \/>\n<meta name=\"author\" content=\"mathemerize\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"mathemerize\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\"},\"author\":{\"name\":\"mathemerize\",\"@id\":\"https:\/\/mathemerize.com\/#\/schema\/person\/104c8bc54f90618130a6665299bc55df\"},\"headline\":\"Prove that \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(sinx)dx = \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(cosx)dx = -\\\\(\\\\pi\\\\over 2\\\\)log 2.\",\"datePublished\":\"2021-10-21T12:26:07+00:00\",\"dateModified\":\"2021-11-01T14:02:40+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\"},\"wordCount\":214,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/mathemerize.com\/#organization\"},\"articleSection\":[\"Integration Questions\",\"Maths Questions\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\",\"url\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\",\"name\":\"Prove that \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(sinx)dx = \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(cosx)dx = -\\\\(\\\\pi\\\\over 2\\\\)log 2.\",\"isPartOf\":{\"@id\":\"https:\/\/mathemerize.com\/#website\"},\"datePublished\":\"2021-10-21T12:26:07+00:00\",\"dateModified\":\"2021-11-01T14:02:40+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mathemerize.com\/prove-that-int_0pi-2-logsinxdx-int_0pi-2-logcosxdx-piover-2log-2\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mathemerize.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Prove that \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(sinx)dx = \\\\(\\\\int_{0}^{\\\\pi\/2}\\\\) log(cosx)dx = -\\\\(\\\\pi\\\\over 2\\\\)log 2.\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mathemerize.com\/#website\",\"url\":\"https:\/\/mathemerize.com\/\",\"name\":\"Mathemerize\",\"description\":\"Maths Tutorials - 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