{"id":7914,"date":"2021-11-10T17:14:38","date_gmt":"2021-11-10T11:44:38","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7914"},"modified":"2022-01-04T19:21:11","modified_gmt":"2022-01-04T13:51:11","slug":"what-is-the-integration-of-log-cos-x-dx","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-integration-of-log-cos-x-dx\/","title":{"rendered":"What is the integration of log cos x dx ?"},"content":{"rendered":"
We have, I = \\(\\int\\) log cos x dx<\/p>\n
By using integraton by parts<\/a>,<\/p>\n I = \\(\\int\\) 1.log cos x dx<\/p>\n Taking log cos x as first function and 1 as second function. Then,<\/p>\n I = log cos x \\(\\int\\) 1 dx – \\(\\int\\) { \\({d\\over dx}\\) (log cos x) \\(\\int\\) 1 dx } dx<\/p>\n I = x log cos x – \\(\\int\\) { \\({-sinx\\over cos x} x\\) } dx<\/p>\n I = x log cos x + \\(\\int\\) x tan x dx<\/p>\n Again using integration by parts<\/a>,<\/p>\n I = x log cos x + x log |sec x | – \\(\\int\\) {1.(log sec x)} dx<\/p>\n I = x log cos x + x log sec x – I<\/p>\n 2I = x log cos x + x log sec x + C<\/p>\n I = \\(x log cos x\\over 2\\) + \\(x log sec x\\over 2\\) + C<\/p>\n Hence, the integration of log cos x with respect to x is \\(x log cos x\\over 2\\) + \\(x log sec x\\over 2\\) + C<\/p>\n What is the integration of \\((log x)^2\\) dx ?<\/a><\/p>\n What is the integration of log 1\/x ?<\/a><\/p>\n What is the integration of 1\/x log x ?<\/a><\/p>\n
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