{"id":5841,"date":"2021-10-03T22:54:52","date_gmt":"2021-10-03T17:24:52","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5841"},"modified":"2021-11-25T23:04:06","modified_gmt":"2021-11-25T17:34:06","slug":"integration-of-cot-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-of-cot-inverse-x\/","title":{"rendered":"Integration of Cot Inverse x"},"content":{"rendered":"
Here you will learn proof of integration of cot inverse x or arccot x and examples based on it.<\/p>\n
Let’s begin –<\/p>\n
\nThe integration of cot inverse x or arccot x is \\(xcot^{-1}x\\) + \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(cot^{-1}x\\) = \\(xcot^{-1}x\\) – \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/p>\n<\/blockquote>\n
Proof : <\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(cot^{-1}x\\) dx<\/p>\n
Let \\(cot^{-1}x\\) = t,<\/p>\n
Then, x = cot t<\/p>\n
\\(\\implies\\) dx = d(cot t) = \\(-cosec^2 t\\) dt<\/p>\n
\\(\\therefore\\) I = \\(\\int\\) \\(cot^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) t \\(-cosec^2 t\\) dt<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = t cot t – \\(\\int\\) 1. (cot t) dt<\/p>\n
I = t cot t – log |sin t| + C<\/p>\n
Since cot t = x \\(\\implies\\) cosec t = \\(\\sqrt{1 + cot^2 t}\\) = \\(\\sqrt{1 + x^2}\\)<\/p>\n
Hence, sin t = \\(1\\over \\sqrt{1 + x^2}\\)<\/p>\n
Now, Put t = \\(cot^{-1}x\\) here,<\/p>\n
\\(\\implies\\) I = x \\(cot^{-1}x\\) – \\(log |{1\\over \\sqrt{ 1+ x^2}}|\\) + C<\/p>\n
Hence, \\(\\int\\) \\(cot^{-1}x\\) dx = \\(xcot^{-1}x\\) + \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Evaluate \\(\\int\\) \\(x cot^{-1} x\\) dx<\/p>\n
Solution<\/strong><\/span> : We have,<\/p>\n
I = \\(\\int\\) \\(x cot^{-1} x\\) dx<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = \\(cot^{-1} x\\) \\(x^2\\over 2\\) – \\(\\int\\) \\(-1\\over 1 + x^2\\) \\(\\times\\) \\(x^2\\over 2\\) dx<\/p>\n
I = \\(tan^{-1} x\\) \\(x^2\\over 2\\) + \\(1\\over 2\\) \\(\\int\\) \\(x^2 + 1 – 1\\over 1 + x^2\\)dx<\/p>\n
= \\(x^2\\over 2\\) \\(cot^{-1} x\\) + \\(1\\over 2\\) \\(\\int\\) 1 – \\(1\\over 1 + x^2\\)dx<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(cot^{-1} x\\) + \\(1\\over 2\\) (x – \\(tan^{-1} x\\)) + C<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(cot^{-1} x\\) + \\(x\\over 2\\) – \\(tan^{-1} x\\over 2\\) + C<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of cot inverse x ?<\/a><\/p>\n
What is the Integration of Cotx ?<\/a><\/p>\n
What is the integration of sec inverse root x ?<\/a><\/p>\n\n\n