Đặt x=2t, dx=2dt
\(2sinx+5cosx+3=2sin2t+5cos2t+3\\ =4sint\cdot cost+5\left(cos^2t-sin^2t\right)+3\left(sin^2t+cos^2t\right)\\ =-2sin^2t+4sint\cdot cost+8cos^2t\)
Ta có:
\(I=\int\frac{2dt}{-2sin^2t+4sint\cdot cost+8cos^2t}\\ =\int\frac{\frac{dt}{cos^2t}}{-tan^2t+2tant+4}=\int\frac{d\left(tant\right)}{-tan^2t+2tant+4}\\ =\int\frac{-d\left(tant\right)}{\left(tant-1+\sqrt{5}\right)\left(tant-1-\sqrt{5}\right)}\\ =\frac{1}{2\sqrt{5}}\int\left(\frac{1}{tant-1+\sqrt{5}}-\frac{1}{tant-1-\sqrt{5}}\right)dt\)
\(=\frac{1}{2\sqrt{5}}ln\left|\frac{tant-1+\sqrt{5}}{tant-1+\sqrt{5}}\right|+C\)
....
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