{"id":5845,"date":"2021-10-03T21:30:10","date_gmt":"2021-10-03T16:00:10","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5845"},"modified":"2021-11-25T23:14:14","modified_gmt":"2021-11-25T17:44:14","slug":"integration-of-cos-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-of-cos-inverse-x\/","title":{"rendered":"Integration of Cos Inverse x"},"content":{"rendered":"
Here you will learn proof of integration of cos inverse x or arccos x and examples based on it.<\/p>\n
Let’s begin –<\/p>\n
\nThe integration of cos inverse x or arccos x is \\(xcos^{-1}x\\) – \\(\\sqrt{1 – x^2}\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(cos^{-1}x\\) = \\(xcos^{-1}x\\) – \\(\\sqrt{1 – x^2}\\) + C<\/p>\n<\/blockquote>\n
Proof :\u00a0<\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(cos^{-1}x\\) dx<\/p>\n
Let \\(cos^{-1}x\\) = t,<\/p>\n
Then, x = cos t<\/p>\n
\\(\\implies\\) dx = d(cos t) = -sin t dt<\/p>\n
\\(\\therefore\\) I = \\(\\int\\) \\(cos^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) -t sint dt<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = t cos t + \\(\\int\\) 1. (-cos t) dt<\/p>\n
I = t cos t – \\(\\int\\) cost dt<\/p>\n
= t cos t – sin t + C<\/p>\n
= t cos t – \\(\\sqrt{1 – cos^2 t}\\) + C<\/p>\n
Now, Put t = \\(cos^{-1}x\\) here<\/p>\n
\\(\\implies\\) I = x \\(cos^{-1}x\\) – \\(\\sqrt{1 – x^2}\\) + C<\/p>\n
Hence, \\(\\int\\) \\(cos^{-1}x\\) dx = x \\(cos^{-1}x\\) – \\(\\sqrt{1 – x^2}\\) + C<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Evaluate \\(\\int\\) \\(x cos^{-1} x\\) dx<\/p>\n
Solution<\/strong><\/span> : We have,<\/p>\n
I = \\(\\int\\)\u00a0 \\(x cos^{-1} x\\) dx<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = \\(cos^{-1} x\\) \\(x^2\\over 2\\) – \\(\\int\\) \\(-1\\over \\sqrt{1 – x^2}\\) \\(\\times\\) \\(x^2\\over 2\\) dx<\/p>\n
I =\u00a0 \\(x^2\\over 2\\) \\(cos^{-1} x\\) – \\(1\\over 2\\) \\(\\int\\) \\(-x^2\\over \\sqrt{1 – x^2}\\) dx<\/p>\n
= \\(x^2\\over 2\\) \\(cos^{-1} x\\) – \\(1\\over 2\\) \\(\\int\\) \\(1 – x^2 – 1\\over \\sqrt{1 – x^2}\\)\u00a0 dx<\/p>\n
= \\(x^2\\over 2\\) \\(cos^{-1} x\\) – \\(1\\over 2\\) { \\(\\int\\) \\(1 – x^2\\over \\sqrt{1 – x^2}\\) – \\(\\int\\) \\(1\\over \\sqrt{1 -x^2}\\) } dx<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(cos^{-1} x\\) – \\(1\\over 2\\) { \\(\\int\\) \\(\\sqrt{1 – x^2}\\) – \\(\\int\\) \\(1\\over \\sqrt{1 -x^2}\\) } dx<\/p>\n
By using integration formula<\/a> of \\(\\sqrt{a^2 – x^2}\\),<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(cos^{-1} x\\) – \\(1\\over 2\\) [{ \\(1\\over 2\\) \\(x\\sqrt{1 – x^2}\\) – \\(1\\over 2\\) \\(sin^{-1} x\\) } – \\(sin ^{-1} x\\) ] + C<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(cos^{-1} x\\) –\u00a0 \\(1\\over 4\\) \\(x\\sqrt{1 – x^2}\\) + \\(3\\over 4\\) \\(sin^{-1} x\\) + C<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of cos inverse x ?<\/a><\/p>\n
What is the Integration of cos x ?<\/a><\/p>\n
What is the integration of cos inverse root x ?<\/a><\/p>\n\n\n