Tích phân \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x\) bằng
\(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\frac{2}{5}\). \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\frac{27}{23}\). \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\frac{34}{27}\). \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\frac{35}{29}\). Hướng dẫn giải:\(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\int\limits^{\frac{\pi}{2}}_0\frac{2\sin x.\cos x+\sin x}{\sqrt{1+3\cos x}}\text{d}x=\int\limits^{\frac{\pi}{2}}_0\frac{\left(2\cos x+1\right)\sin x\text{d}x}{\sqrt{1+3\cos x}}\)
Đặt \(t=\sqrt{1+3\cos x}\) => \(t^2=1+3\cos x\)
⇒ \(2t\text{d}t=-3\sin x\text{d}x\) và \(\cos x=\frac{t^2-1}{3}\)
Thay vào ta được tích phân đa thức của t:
\(\int\limits^1_2\frac{\left(2.\frac{t^2-1}{3}+1\right).\left(-\frac{2}{3}\right)t\text{d}t}{t}\)
\(=\int\limits^2_1\frac{2}{9}\left(2t^2+1\right)\text{d}t\)
\(=\frac{2}{9}\left(2.\frac{t^3}{3}+t\right)|^2_1\)
\(=\frac{34}{29}\).