Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\cos x}{\left(2+\sin x\right)\left(1+\sin x\right)}dx$ equals

(a) $\log \left(\frac{2}{3}\right)$

(b) $\log \left(\frac{3}{2}\right)$

(c) $\log \left(\frac{3}{4}\right)$

(d) $\log \left(\frac{4}{3}\right)$

Answer 1

(d) $\log \left(\frac{4}{3}\right)$

$$\begin{gathered} \mathrm{Let},I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{\left(2+\sin x\right)\left(1+\sin x\right)}dx \\ \mathrm{Let}\sin x,\mathrm{then}\cos xdx=dt \\ \mathrm{When}x=0,t=0,x=\frac{\mathrm{\pi }}{2},t=1 \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ I={\int }_0^1\frac{\mathrm{dt}}{\left(2+\mathrm{t}\right)\left(1+\mathrm{t}\right)} \\ ={\int }_0^1\left[\frac{-1}{2+\mathrm{t}}+\frac{1}{1+\mathrm{t}}\right]\mathrm{dt} \\ ={\left[-\log \left(2+\mathrm{t}\right)+\log \left(1+\mathrm{t}\right)\right]}_0^1 \\ ={\left[\log \left(1+\mathrm{t}\right)-\log \left(2+\mathrm{t}\right)\right]}_0^1 \\ =\log 2-\log 3-\log 1+\log 2 \\ =\log \frac{4}{3} \end{gathered}$$