Tích phân \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos x\text{d}x}{1+\cos x}\) bằng
\(-1+\ln2\). \(-1+3\ln2\). \(-1+2\ln2\). \(2+2\ln2\). Hướng dẫn giải:\(I=\) \(\int\limits^{\frac{\pi}{2}}_0\frac{\sin2x\cos x\text{d}x}{1+\cos x}=\int\limits^{\frac{\pi}{2}}_0\frac{2\sin x\cos x.\cos x}{1+\cos x}\text{d}x\)
\(=\int\limits^{\frac{\pi}{2}}_0\frac{2\cos^2x}{1+\cos x}\sin x\text{d}x\)
Đặt \(t=1+\cos x\) \(\Rightarrow\text{d}t=-\sin x\text{d}x\) ; \(\cos^2x=\left(1-t\right)^2\)
Đổi cận \(x|^{\frac{\pi}{2}}_0\Rightarrow t|^1_2\), tích phân đã cho bằng:
\(I=-\int\limits^1_2\frac{2\left(1-t\right)^2}{t}\text{d}t\)
\(=2\int\limits^2_1\frac{1-2t+t^2}{t}\text{d}t\)
\(=2\int\limits^2_1\left(\frac{1}{t}-2+t\right)\text{d}t\)
\(=2\left(\ln t-2t+\frac{t^2}{2}\right)|^2_1\)
\(=2\left(\ln2-4+2+2-\frac{1}{2}\right)\)
\(=2\ln2-1\).