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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 $I = \frac{4}{{25}}\left( {1 + {e^{\frac{3}{4}}}} \right)$
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Đăng bài 09-05-12 09:11 AM
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