Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\)

Solution :

I = \(\int\) \(dx\over {3sinx + 4cosx}\) = \(\int\) \(dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}\) = \(\int\) \(sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}\)

let \(tan{x\over 2}\) = t,

\(\therefore\)  \({1\over 2}sec^2{x\over 2}\)dx = dt

so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\) \(\int\) \(dt\over {{25\over 16}-{(t-{3\over 4})}^2}\)

= \(1\over 2\) \(1\over {2({5\over 4})}\) \(ln|{{{5\over 4}+(t-{3\over 4})}\over {{5\over 4}-(t-{3\over 4})}}|\) + C = \(1\over 5\) \(ln|{1+2tan{x\over 2}\over {4-2tan{x\over 2}}}|\) + C


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