Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sin x\cos x}{{\cos }^{2}x+3\cos x+2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x\cos x}{{\cos }^{2}x+3\cos x+2}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ \mathrm{Let}\cos x=t.\mathrm{Then},-\sin xdx=dt \\ \mathrm{When}x=0,t=1\mathrm{and}x=\frac{\mathrm{\pi }}{2},t=0 \\ \therefore I=-{\int }_1^0\frac{tdt}{t^2+3t+2} \\ \Rightarrow I={\int }_1^0\frac{-tdt}{\left(t+{\displaystyle 2}\right)\left(t+1\right)} \\ \Rightarrow I={\int }_1^0\left(\frac{1}{\left(t+1\right)}-\frac{2}{\left(t+2\right)}\right)dt \\ \Rightarrow I={\left[\log \left(t+1\right)-2\log \left(t+2\right)\right]}_1^0 \\ \Rightarrow I={\left[\log \frac{\left(t+1\right)}{{\left(t+2\right)}^2}\right]}_0^1 \\ \Rightarrow I={\left[\log \left(\frac{1}{4}\right)-\log \left(\frac{2}{9}\right)\right]}_0^1 \\ \Rightarrow I=\log \frac{9}{8} \end{gathered}$$