Đáp án:
\[ - \frac{1}{4}\left( {2x + 3} \right)\cos 4x + \frac{1}{8}\sin 4x + C\]
Giải thích các bước giải:
Ta đặt:
\(\begin{array}{l}
\left\{ \begin{array}{l}
u = 2x + 3\\
v' = \sin 4x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = 2\\
v = - \frac{1}{4}\cos 4x
\end{array} \right.\\
\int\limits_0^{\frac{\pi }{4}} {\left( {2x + 3} \right).\sin 4xdx} = - \frac{1}{4}\cos 4x.\left( {2x + 3} \right) - \int\limits_0^{\frac{\pi }{4}} {2.\left( { - \frac{1}{4}\cos 4x} \right)dx} \\
= - \frac{1}{4}\left( {2x + 3} \right)\cos 4x + \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 4xdx} \\
= - \frac{1}{4}\left( {2x + 3} \right)\cos 4x + \frac{1}{8}\sin 4x + C
\end{array}\)
