{"id":5839,"date":"2021-10-03T22:55:21","date_gmt":"2021-10-03T17:25:21","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5839"},"modified":"2021-11-25T23:16:21","modified_gmt":"2021-11-25T17:46:21","slug":"integration-of-sec-inverse-x-and-cosec-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-of-sec-inverse-x-and-cosec-inverse-x\/","title":{"rendered":"Integration of Sec Inverse x and Cosec Inverse x"},"content":{"rendered":"
Here you will learn proof of integration of sec inverse x\u00a0 and cosec inverse x.<\/p>\n
Let’s begin –<\/p>\n
\nThe integration of sec inverse x or arcsec x is \\(xsec^{-1}x\\) – \\(log|x + \\sqrt{x^2 – 1}|\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(sec^{-1}x\\) = \\(xsec^{-1}x\\) – \\(log|x + \\sqrt{x^2 – 1}|\\) + C<\/p>\n<\/blockquote>\n
Proof :\u00a0<\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(sec^{-1}x\\) dx<\/p>\n
Let \\(sec^{-1}x\\) = t,<\/p>\n
Then, x = sec t<\/p>\n
\\(\\implies\\) dx = d(sec t) = sec t tan t dt<\/p>\n
\\(\\therefore\\) I = \\(\\int\\) \\(sec^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) t (sec t tan t) dt<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = t sec t – \\(\\int\\) 1. (sec t) dt<\/p>\n
I = t sec t – log |sec t + tan t| + C<\/p>\n
Since tan t = \\(\\sqrt{sec^2 t – 1}\\)\u00a0<\/p>\n
\\(\\implies\\) I = t sec t – \\(log |sec t + \\sqrt{sec^2t – 1}|\\) + C<\/p>\n
Now, Put t = \\(sec^{-1}x\\) here,<\/p>\n
\\(\\implies\\) I = x \\(sec^{-1}x\\) – \\(log|x + \\sqrt{x^2 – 1}|\\) + C<\/p>\n
Hence, \\(\\int\\) \\(sec^{-1}x\\) dx = x \\(sec^{-1}x\\) – \\(log|x + \\sqrt{x^2 – 1}|\\) + C<\/p>\n<\/blockquote>\n
Integration of Cosec Inverse x<\/strong><\/h2>\n
\nThe integration of cosec inverse x or arccosec x is x\\(cosec^{-1}x\\) – \\(log|x – \\sqrt{x^2 – 1}|\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(cosec^{-1}x\\) = x\\(cosec^{-1}x\\) – \\(log|x – \\sqrt{x^2 – 1}|\\) + C<\/p>\n<\/blockquote>\n
Proof :\u00a0<\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(cosec^{-1}x\\) dx<\/p>\n
Let \\(cosec^{-1}x\\) = t,<\/p>\n
Then, x = cosec t<\/p>\n
\\(\\implies\\) dx = d(cosec t) = -cosec t cot t dt<\/p>\n
\\(\\therefore\\) I = \\(\\int\\) \\(cosec^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) t (-cosec t cot t) dt<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = t cosec t – \\(\\int\\) 1. (cosec t) dt<\/p>\n
I = t cosec t – log |cosec – cot t| + C<\/p>\n
Since cot t = \\(\\sqrt{cosec^2 t – 1}\\)\u00a0<\/p>\n
\\(\\implies\\) I = t cosec t – \\(log |cosec t – \\sqrt{cosec^2t – 1}|\\) + C<\/p>\n
Now, Put t = \\(cosec^{-1}x\\) here,<\/p>\n
\\(\\implies\\) I = x \\(cosec^{-1}x\\) – \\(log|x – \\sqrt{x^2 – 1}|\\) + C<\/p>\n
Hence, \\(\\int\\) \\(cosec^{-1}x\\) dx = x \\(cosec^{-1}x\\) – \\(log|x – \\sqrt{x^2 – 1}|\\) + C<\/p>\n<\/blockquote>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of cosec inverse x ?<\/a><\/p>\n
What is the Differentiation of sec inverse x ?<\/a><\/p>\n
What is the integration of sec inverse root x ?<\/a><\/p>\n\n\n