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
Evaluate : \\(\\int\\) \\(cos^4xdx\\over {sin^3x{(sin^5x + cos^5x)^{3\\over 5}}}\\)<\/a><\/p>\n
Evaluate : \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\)<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[52,43],"tags":[],"yoast_head":"\n