tính tích phân

Question

Tính : $I = \int_1^{{e^\pi }} {\cos \left( {\ln x} \right)dx} $

score 2 · Answer 1

Đặt $u =cos(lnx)\Rightarrow du=-\frac{sin(lnx)dx}{x}$
      $dv =dx\Rightarrow v=x$
Suy ra : $I = x\cos \left( {\ln x} \right)\left| {_1^{{e^\pi }}} \right. + \int_1^{{e^\pi }} {\sin \left( {\ln x} \right)dx} = - {e^\pi } - 1 + K$ với $K = \int_1^{{e^\pi }} {\sin \left( {\ln x} \right)dx} $
Tương tự : để tính $K$ ta đặt : $u=sin(lnx)\Rightarrow du=\frac{cosx(lnx)dx}{x}$
                                                 $dv=dx\Rightarrow v=x$
Suy ra : $K = x\sin \left( {\ln x} \right)\left| {_1^{{e^\pi }}} \right. - I = 0 - I = - I$
Từ đó : $I =-e^x-1-1\Rightarrow I=-\frac{1}{2}(e^x+1)$

Ad hoạt động chăm quá – babysexy156 14-07-12 08:17 AM

Cám ơn bạn nhé ! – anh_chang_zaizai_90 13-07-12 12:57 PM

score 3 · Answer 2

Đặt $\ln x=t\Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=e^tdt$.
Đổi cận: $x=1\;\Rightarrow t=0$
                $x=e^\pi\Rightarrow t=\pi$
Ta có: $I=\int\limits_0^\pi e^t\cos tdt=\int\limits_0^\pi\cos td(e^t)$
                                            $=e^t\cos t\left|\begin{array}{I}\pi\\0\end{array}\right.-\int\limits_0^\pi e^td(\cos t)$
                                            $=-e^\pi-1+\int\limits_0^\pi e^t\sin tdt$
                                            $=-e^\pi-1+\int\limits_0^\pi \sin td(e^t)$
                                            $=-e^\pi-1+(e^t\sin t)\left|\begin{array}{I}\pi\\0\end{array}\right.-\int\limits_0^\pi e^td(\sin t)$
                                            $=-e^\pi-1-\int\limits_0^\pi e^t\cos tdt=-e^\pi-1-I$
             $\Rightarrow I=\frac{-e^\pi-1}{2}$