Ai làm giúp e với ạ. E đang cần gấp ạ Nguyên hàm của e^2x . Cos (1-3x) dx

Đáp án:

\[\int {{e^{2x}}.\cos \left( {1 - 3x} \right)dx}  = \frac{2}{{13}}{e^{2x}}.\left[ {\cos \left( {1 - 3x} \right) - \frac{3}{2}\sin \left( {1 - 3x} \right)} \right]\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
u = \cos \left( {1 - 3x} \right)\\
v' = {e^{2x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \left( {1 - 3x} \right).\left( { - \sin \left( {1 - 3x} \right)} \right) = 3\sin \left( {1 - 3x} \right)\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\\
 \Rightarrow I = \int {{e^{2x}}.\cos \left( {1 - 3x} \right)dx} \\
 = \frac{1}{2}{e^{2x}}.\cos \left( {1 - 3x} \right) - \int {3\sin \left( {1 - 3x} \right).\frac{1}{2}{e^{2x}}dx} \\
 = \frac{1}{2}{e^{2x}}.\cos \left( {1 - 3x} \right) - \frac{3}{2}\int {\sin \left( {1 - 3x} \right){e^{2x}}dx} \\
\left\{ \begin{array}{l}
u = \sin \left( {1 - 3x} \right)\\
v' = {e^{2x}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
u' = \left( {1 - 3x} \right)'.cos\left( {1 - 3x} \right) =  - 3\cos \left( {1 - 3x} \right)\\
v = \frac{1}{2}{e^{2x}}
\end{array} \right.\\
 \Rightarrow {I_1} = \int {\sin \left( {1 - 3x} \right){e^{2x}}dx} \\
 = \frac{1}{2}{e^{2x}}.\sin \left( {1 - 3x} \right) - \int {\left( { - 3} \right).\cos \left( {1 - 3x} \right).\frac{1}{2}{e^{2x}}dx} \\
 = \frac{1}{2}{e^{2x}}\sin \left( {1 - 3x} \right) + \frac{3}{2}\int {{e^{2x}}.\cos \left( {1 - 3x} \right)dx} \\
I = \frac{1}{2}{e^{2x}}.\cos \left( {1 - 3x} \right) - \frac{3}{2}{I_1}\\
 = \frac{1}{2}{e^{2x}}.\cos \left( {1 - 3x} \right) - \frac{3}{2}.\left( {\frac{1}{2}{e^{2x}}\sin \left( {1 - 3x} \right) + \frac{3}{2}\int {{e^{2x}}.\cos \left( {1 - 3x} \right)dx} } \right)\\
 = \frac{1}{2}{e^{2x}}.\left[ {\cos \left( {1 - 3x} \right) - \frac{3}{2}\sin \left( {1 - 3x} \right)} \right] - \frac{9}{4}\int {{e^{2x}}.\cos \left( {1 - 3x} \right)dx} \\
 = \frac{1}{2}{e^{2x}}.\left[ {\cos \left( {1 - 3x} \right) - \frac{3}{2}\sin \left( {1 - 3x} \right)} \right] - \frac{9}{4}I\\
 \Rightarrow I = \frac{2}{{13}}{e^{2x}}.\left[ {\cos \left( {1 - 3x} \right) - \frac{3}{2}\sin \left( {1 - 3x} \right)} \right]
\end{array}\)