{"id":5833,"date":"2021-10-03T22:10:18","date_gmt":"2021-10-03T16:40:18","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5833"},"modified":"2021-11-25T23:13:35","modified_gmt":"2021-11-25T17:43:35","slug":"integration-of-tan-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-of-tan-inverse-x\/","title":{"rendered":"Integration of Tan Inverse x"},"content":{"rendered":"
Here you will learn proof of integration of tan inverse x or arctan x and examples based on it.<\/p>\n
Let’s begin –<\/p>\n
\nThe integration of tan inverse x or arctan x is \\(xtan^{-1}x\\) – \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(tan^{-1}x\\) = \\(xtan^{-1}x\\) – \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/p>\n<\/blockquote>\n
Proof :\u00a0<\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(tan^{-1}x\\) dx<\/p>\n
Let \\(tan^{-1}x\\) = t,<\/p>\n
Then, x = tan t<\/p>\n
\\(\\implies\\) dx = d(tan t) = \\(sec^2 t\\) dt<\/p>\n
\\(\\therefore\\) I = \\(\\int\\) \\(tan^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) t \\(sec^2 t\\) dt<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = t tan t – \\(\\int\\) 1. (tan t) dt<\/p>\n
I = t tan t + log |cos t| + C<\/p>\n
Since tan t = x \\(\\implies\\) cost = \\(1\\over \\sqrt{1 + tan^2 t}\\) = \\(1\\over \\sqrt{1 + x^2}\\)<\/p>\n
Now, Put t = \\(tan^{-1}x\\) here,<\/p>\n
\\(\\implies\\) I = x \\(tan^{-1}x\\) + \\(log |{1\\over \\sqrt{ 1+ x^2}}|\\) + C<\/p>\n
Hence, \\(\\int\\) \\(tan^{-1}x\\) dx = \\(xtan^{-1}x\\) – \\(1\\over 2\\) \\(log |1 + x^2|\\) + C<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Evaluate \\(\\int\\) \\(x tan^{-1} x\\) dx<\/p>\n
Solution<\/strong><\/span> : We have,<\/p>\n
I = \\(\\int\\)\u00a0 \\(x tan^{-1} x\\) dx<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = \\(tan^{-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\\) \\(tan^{-1} x\\) – \\(1\\over 2\\) \\(\\int\\)\u00a0 1 – \\(1\\over 1 + x^2\\)dx<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(tan^{-1} x\\) – \\(1\\over 2\\) (x\u00a0 – \\(tan^{-1} x\\)) + C<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(tan^{-1} x\\) – \\(x\\over 2\\) + \\(tan^{-1} x\\over 2\\) + C<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of tan inverse x ?<\/a><\/p>\n
What is the Integration of tan x ?<\/a><\/p>\n
What is the integration of tan inverse root x ?<\/a><\/p>\n\n\n