I=$\int\limits_{0}^{\pi/2}$$e^{2}$$\sin3 x$dx

đầu tiên tính như thế này
$\begin{cases}u= sin3x\\ dv=e^{2x} \end{cases}$ =$\begin{cases}du=3cos3xdx \\ v= \frac{1}{2}e^{2}\end{cases}$


$\frac{1}{2} $$e^{2x}$sin3x$\begin{matrix} \prod_{0}^{\frac{\pi}{2}}   &  \end{matrix}$$\int\limits_{0}^{\frac{\pi}{2}}$$\frac{1}{2}$$e^{2x}$3cos3xdx
tới khúc này mình hết hiểu rồi mong các bạn chỉ giúp mình với
I+ 9I/4=$\frac{3-2e^{\pi}}{4}$ 
I=$\frac{3-2e^{\pi}}{13} $
$\begin{cases}u= sin3x\\ dv=e^{2x}dx \end{cases}$ =$\begin{cases}du=3cos3xdx \\ v= \frac{1}{2}e^{2x}\end{cases}$
$I=\frac{1}{2} e^{2x}sin3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix} - \int\limits_{0}^{\frac{\pi}{2}}\frac{1}{2}e^{2x}3cos3xdx$
$I_1=\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{2}e^{2x}3cos3xdx=\frac{3}{2}\int\limits_{0}^{\frac{\pi}{2}}e^{2x}cos3xdx$
$\begin{cases}u=cos3x\\ dv=e^{2x}dx \end{cases}$ =$\begin{cases}du=-3sin3xdx \\ v= \frac{1}{2}e^{2x}\end{cases}$
$I_1=\frac{1}{2} e^{2x}cos3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix} + \int\limits_{0}^{\frac{\pi}{2}}\frac{1}{2}e^{2x}3sin3xdx$
$I_1=\frac{1}{2} e^{2x}cos3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix} + \frac{3}{2}I$
$\Rightarrow I=\frac{1}{2} e^{2x}sin3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix}-\frac{1}{2} e^{2x}cos3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix}-\frac{3}{2}I$
$\Rightarrow \frac{5}{2}I=\frac{1}{2} e^{2x}sin3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix}-\frac{1}{2} e^{2x}cos3x|\begin{matrix}\frac{\Pi}{2}\\ 0 \end{matrix}$

Tại vì bạn chưa xem lại đề bài xem I là gì –  gunbk92 01-03-13 07:10 PM
tại sao lại xuất hiện 3/2I ở I1 nhỉ? –  Huynh ICT 01-03-13 07:07 PM
Hãy ấn chữ V dưới chữ đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  gunbk92 28-02-13 12:32 AM

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