{"id":7786,"date":"2021-11-06T14:05:54","date_gmt":"2021-11-06T08:35:54","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7786"},"modified":"2021-11-06T15:35:39","modified_gmt":"2021-11-06T10:05:39","slug":"what-is-the-integration-of-cos-inverse-root-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-integration-of-cos-inverse-root-x\/","title":{"rendered":"What is the integration of cos inverse root x ?"},"content":{"rendered":"

Solution :<\/h2>\n

We have, I = \\(cos^{-1}\\sqrt{x}\\) . 1 dx<\/p>\n

By Applying integration by parts,<\/p>\n

Taking \\(cos^{-1}\\sqrt{x}\\) as first function and 1 as second function. Then<\/p>\n

I = \\(cos^{-1}\\sqrt{x}\\) \\(\\int\\) 1 dx – \\(\\int\\) {\\(d\\over dx\\)\\(cos^{-1}\\sqrt{x}\\) \\(\\int\\) 1 dx } dx<\/p>\n

I = x\\(cos^{-1}\\sqrt{x}\\) – \\(\\int\\) \\(-1\\over 2\\sqrt{(1-x)}\\sqrt{x}\\) . x dx<\/p>\n

I = x\\(cos^{-1}\\sqrt{x}\\) – \\(\\int\\) \\(-1\\over 2\\) \\(\\sqrt{x}\\over \\sqrt{(1-x)}\\) dx<\/p>\n

Put x = \\(sin^2 t\\)<\/p>\n

dx = 2 sin t cos t dt<\/p>\n

\\(\\implies\\) I = x\\(cos^{-1}\\sqrt{x}\\) + { \\(1\\over 2\\) \\(\\int\\) \\(2cos^2 t\\) dt }<\/p>\n

\\(\\implies\\) I = x\\(cos^{-1}\\sqrt{x}\\)\u00a0 + { \\(1\\over 2\\) \\(\\int\\) (1 – cos 2t) dt }<\/p>\n

\\(\\implies\\) I = x\\(cos^{-1}\\sqrt{x}\\) + \\(1\\over 2\\)t – \\(1\\over 4\\)\\(sin 2t\\) + C<\/p>\n

Now, sin 2t = 2 sin t cos t = \\(2\\sqrt{x} \\sqrt{1 – x}\\) = \\(2\\sqrt{x – x^2}\\)<\/p>\n

I = x\\(cos^{-1}\\sqrt{x}\\) + \\(1\\over 2\\)\\(cos^{-1}\\sqrt{x}\\) – \\(1\\over 4\\) (\\(2\\sqrt{x – x^2}\\))\u00a0 + C<\/p>\n

I = (x + \\({1\\over 2}\\))\\(cos^{-1}\\sqrt{x}\\) – \\(1\\over 2\\) \\(\\sqrt{x – x^2}\\) \u00a0+ C<\/p>\n

\n

Similar Questions<\/h3>\n

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

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

What is integration of sin inverse cos x ?<\/a><\/p>\n

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

What is the integration of sin inverse x whole square ?<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have, I = \\(cos^{-1}\\sqrt{x}\\) . 1 dx By Applying integration by parts, Taking \\(cos^{-1}\\sqrt{x}\\) as first function and 1 as second function. Then I = \\(cos^{-1}\\sqrt{x}\\) \\(\\int\\) 1 dx – \\(\\int\\) {\\(d\\over dx\\)\\(cos^{-1}\\sqrt{x}\\) \\(\\int\\) 1 dx } dx I = x\\(cos^{-1}\\sqrt{x}\\) – \\(\\int\\) \\(-1\\over 2\\sqrt{(1-x)}\\sqrt{x}\\) . x dx I = x\\(cos^{-1}\\sqrt{x}\\) – …<\/p>\n

What is the integration of cos inverse root x ?<\/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":"","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":"\nWhat is the integration of cos inverse root x ? - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn what is the integration of cos inverse root x by using integration by parts.\" \/>\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\/what-is-the-integration-of-cos-inverse-root-x\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is the integration of cos inverse root x ? 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