|
|
Đặt $u = {e^{3x\,\,\,\,}}\,\,\,\,\, \Rightarrow \,\,\,du\, = \,\,3{e^{3x\,}}dx$
$dv = \sin \,4x\,dx\,\,\,\, \Rightarrow \,\,\,v = \,\, - \frac{1}{4}co\,sx\,$
$\begin{array}{l}
I
= \,\,\,\, - \left. {\frac{1}{4}{e^{3x\,}}cosx} \right]_0^{\frac{\pi
}{4}} + \frac{3}{4}\int\limits_0^{\frac{\pi }{4}} {{e^{3x\,}}cos4x} \\
\,\,\, = \frac{1}{4}\left( {1 + {e^{\frac{3}{4}}}} \right) + \frac{3}{4}K
\end{array}$
Tính $K$ : Đặt $u = {e^{3x\,\,\,\,}}\,\,\,\,\, \Rightarrow \,\,\,du\, = \,\,3{e^{3x\,}}dx$
$dv = cos4xdx\,\,\,\, \Rightarrow \,\,\,v = \,\,\frac{1}{4}\sin \,4x\,$
$\begin{array}{l}
K = \left. {\frac{1}{4}{e^{3x\,}}\sin \,4x} \right]_0^{\frac{\pi }{4}} - \frac{3}{4}I = - \frac{3}{4}I\\
I
= \frac{1}{4}\left( {1 + {e^{\frac{3}{4}}}} \right) -
\frac{9}{{16}}I\,\,\, \Rightarrow \,\,\frac{{25}}{{16}}I =
\frac{1}{4}\left( {1 + {e^{\frac{3}{4}}}} \right)
\end{array}$
Vậy $I = \frac{4}{{25}}\left( {1 + {e^{\frac{3}{4}}}} \right)$
|
|
|
Trả lời 28-06-12 12:20 AM
|
|