## Calculus: Early Transcendentals (2nd Edition)

$$\ln \left( {\frac{9}{8}} \right)$$
\eqalign{ & \int_0^{\pi /6} {\frac{{\sin 2y}}{{{{\sin }^2}y + 2}}dy} \cr & {\text{use the hint }}\sin 2y = 2\sin y\cos y \cr & = \int_0^{\pi /6} {\frac{{2\sin y\cos y}}{{{{\sin }^2}y + 2}}dy} \cr & {\text{substitute }}u = {\sin ^2}y + 2,{\text{ }}du = 2\sin y\cos ydy \cr & {\text{express the limits in terms of }}u \cr & y = 0{\text{ implies }}u = {\sin ^2}\left( 0 \right) + 2 = 2 \cr & y = \pi /6{\text{ implies }}u = {\sin ^2}\left( {\pi /6} \right) + 2 = 9/4 \cr & {\text{The entire integration is carried out as follows}} \cr & \int_0^{\pi /6} {\frac{{2\sin y\cos y}}{{{{\sin }^2}y + 2}}dy} = \int_2^{9/4} {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\ln \left| u \right|} \right)} \right|_2^{9/4} \cr & {\text{use the fundamental theorem}} \cr & = \ln \left( {9/4} \right) - \ln \left( 2 \right) \cr & {\text{simplify}} \cr & = \ln \left( {\frac{{9/4}}{2}} \right) \cr & = \ln \left( {\frac{9}{8}} \right) \cr}