Tính tích phân sau

$B=\int_{0}^{\frac{\pi }{4}}{\left( 2x+3 \right).\sin 4x.dx}.\text{ }\breve{a}t:\left\{ \begin{align}  & u=2x+3 \\  & dv=\sin 4x.dx \\ \end{align} \right.\,\,$$ta\,\,c\acute{o}:\left\{ \begin{align}  & du=2.dx \\  & v=-\frac{1}{4}\cos 4x.dx \\ \end{align} \right.$

$\Rightarrow B=\left( -\frac{1}{4}\left( 2x+3 \right)\cos 4x \right)_{0}^{\frac{\pi }{4}}+\frac{1}{2}\int_{0}^{\frac{\pi }{4}}{\cos 4x.dx}=\left( -\frac{1}{4}\left( 2x+3 \right)\cos 4x+\frac{1}{2}.\frac{1}{4}.\sin 4x \right)_{0}^{\frac{\pi }{4}}=\frac{\pi }{8}+\frac{3}{2}$