{"id":6807,"date":"2021-10-21T17:49:02","date_gmt":"2021-10-21T12:19:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6807"},"modified":"2021-11-01T19:31:36","modified_gmt":"2021-11-01T14:01:36","slug":"evaluate-int-dxover-3sinx-4cosx","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/evaluate-int-dxover-3sinx-4cosx\/","title":{"rendered":"Evaluate : \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\)"},"content":{"rendered":"
I = \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\) = \\(\\int\\) \\(dx\\over {3[{2tan{x\\over 2}\\over {1+tan^2{x\\over 2}}}] + 4[{1-tan^2{x\\over 2}\\over {1+tan^2{x\\over 2}}}]}\\) = \\(\\int\\) \\(sec^2{x\\over 2}dx\\over {4+6tan{x\\over 2}-4tan^2{x\\over 2}}\\)<\/p>\n
let \\(tan{x\\over 2}\\) = t,<\/p>\n
\\(\\therefore\\)\u00a0 \\({1\\over 2}sec^2{x\\over 2}\\)dx = dt<\/p>\n
so I = \\(\\int\\) \\(2dt\\over {4+6t-4t^2}\\) = \\(1\\over 2\\) \\(\\int\\) \\(dt\\over {1-(t^2-{3\\over 2}t})\\) = \\(1\\over 2\\) \\(\\int\\) \\(dt\\over {{25\\over 16}-{(t-{3\\over 4})}^2}\\)<\/p>\n
= \\(1\\over 2\\) \\(1\\over {2({5\\over 4})}\\) \\(ln|{{{5\\over 4}+(t-{3\\over 4})}\\over {{5\\over 4}-(t-{3\\over 4})}}|\\) + C = \\(1\\over 5\\) \\(ln|{1+2tan{x\\over 2}\\over {4-2tan{x\\over 2}}}|\\) + C<\/p>\n
What is the integration of x tan inverse x dx ?<\/a><\/p>\n Prove that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2.<\/a><\/p>\n Evaluate : \\(\\int\\) \\(cos^4xdx\\over {sin^3x{(sin^5x + cos^5x)^{3\\over 5}}}\\)<\/a><\/p>\n