Biết rằng 4 cos x - 2 sin x / sin x + 3 cos x dx = a pi /2 + b ln 2

Ta có \[\int\limits_0^{\frac{\pi }{4}} {\frac{{4\cos x - 2\sin x}}{{\sin x + 3\cos x}}{\rm{d}}x}  = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + 3\cos x + \cos x - 3\sin x}}{{\sin x + 3\cos x}}{\rm{d}}x}  = \int\limits_0^{\frac{\pi }{4}} {\left( {1 + \frac{{{{\left( {\sin x + 3\cos x} \right)}^{\prime }}}}{{\sin x + 3\cos x}}} \right){\rm{d}}x} \]

\[ = \left. x \right|\mathop {}\limits_0^{\frac{\pi }{4}}  + \left. {\ln \left( {\sin x + 3\cos x} \right)} \right|\mathop {}\limits_0^{\frac{\pi }{4}}  = \frac{\pi }{4} + \ln \left( {2\sqrt 2 } \right) - \ln 3 = \frac{1}{2}.\frac{\pi }{2} + \frac{3}{2}\ln 2 - \ln 3\].

Từ đây ta có \(a = \frac{1}{2},\,b = \frac{3}{2},\,c = 1\) nên \(P = \frac{3}{4}\).