Question

Solve ${\int }_0^{\pi /2}\frac{\sin x\cos x}{{\cos }^{2}x+3\cos x+2}dx$

Answer 1

$letcosx=t\Rightarrow -sinxdx=dt⟶\left(1\right)alsoforx=0,t=1andforx=\frac{\pi }{2},t=0hence,\int \frac{sinxcosx}{{cos}^{2}x+3cosx+2}dx=-\int \frac{t}{t^2+3t+2}dt=-\int \frac{t}{(t+1)(t+2)}dtintegratingbypartialfractions\because \frac{t}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}\Rightarrow t=A(t+2)+B(t+1)\Rightarrow t=(A+B)t+(2A+B)comparingbothsidewegetA+B=1and2A+B=0solvingboththeequationwebgetA=-1andB=2\therefore -{\int }_1^0\frac{t}{(t+1)(t+2)}dt=-{\int }_1^0[\frac{-1}{t+1}+\frac{2}{t+2}]dt={\int }_1^0[\frac{1}{t+1}-\frac{2}{t+2}]dt={\left[log\right(t+1\left)\right]}_1^0-{\left[log\right(t+2\left)\right]}_1^0=log\frac{9}{8}$