* Đặt $\begin{cases}u = \cos ^2 (\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -2 \sin (\ln x)\cos (\ln x) \frac{1}{x}dx = - \sin (2\ln x)\frac{dx}{x} \\ v=x \end{cases} $$ T = ( x \cos ^2 (\ln x)) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + \int\limits_{1}^{e^{\frac{\pi}{2}}} \sin(2 \ln x) dx = -1 + S$ (1)* Đặt $\begin{cases}u = \sin (2 \ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = 2 \cos (2\ln x)\frac{dx}{x} \\ v=x \end{cases}$$ S = ( x \sin (2\ln x)) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. - 2 \int\limits_{1}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -1 \int\limits_{1}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -2R$ (2)* Đặt : $\begin{cases}u = \cos (2\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -\frac{2}{x} \sin (2\ln x)dx \\ v = x \end{cases}$$ R = (xcos (2\ln x))\left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + 2 \int\limits_{1}^{e^{\frac{\pi}{2}}} \sin (2\ln x)dx = -e^{\frac{\pi}{2} } - 1 + 2S$ (3)Thay (3) vào (2) ta có : $S = 2e^{\frac{\pi}{2} }+2-4S \Rightarrow S =\frac{2}{5}(e^{\frac{\pi}{2} } + 1)$ (4)Thay (4) vào (1) ta có : $T = -1 + \frac{2}{5}(e^{\frac{\pi}{2} } + 1) =\frac{1}{5} (2e^{\frac{\pi}{2} }-3).$
* Đặt $\begin{cases}u = \cos ^2 (\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -2 \sin (\ln x)\cos (\ln x) \frac{1}{x}dx = - \sin (2\ln x)\frac{dx}{x} \\ v=x \end{cases} $$ T = ( x \cos ^2 (\ln x) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + \int\limits_{0}^{e^{\frac{\pi}{2}}} \sin(2 \ln x) dx \equiv -1 + S$ (1)* Đặt $\begin{cases}u = \sin (2 \ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = 2 \cos (2\ln x)\frac{dx}{x} \\ v=x \end{cases}$$ S = ( x \sin (2\ln x) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. - 2 \int\limits_{0}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -1 \int\limits_{0}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -2R$ (2)* Đặt : $\begin{cases}u = \cos (2\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -\frac{2}{x} \sin (2\ln x)dx \\ v = x \end{cases}$$ R = (xcos (2\ln x))\left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + 2 \int\limits_{0}^{e^{\frac{\pi}{2}}} \sin (2\ln x)dx = -e^{\frac{\pi}{2} } - 1 + 2S$ (3)Thay (3) vào (2) ta có : $S = 2e^{\frac{\pi}{2} }+2-4S \Rightarrow S =\frac{2}{5}(e^{\frac{\pi}{2} } + 1)$ (4)Thay (4) vào (1) ta có : $T = -1 + \frac{2}{5}(e^{\frac{\pi}{2} } + 1) =\frac{1}{5} (2e^{\frac{\pi}{2} }-3).$
* Đặt $\begin{cases}u = \cos ^2 (\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -2 \sin (\ln x)\cos (\ln x) \frac{1}{x}dx = - \sin (2\ln x)\frac{dx}{x} \\ v=x \end{cases} $$ T = ( x \cos ^2 (\ln x)
) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + \int\limits_{
1}^{e^{\frac{\pi}{2}}} \sin(2 \ln x) dx
= -1 + S$ (1)* Đặt $\begin{cases}u = \sin (2 \ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = 2 \cos (2\ln x)\frac{dx}{x} \\ v=x \end{cases}$$ S = ( x \sin (2\ln x)
) \left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. - 2 \int\limits_{
1}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -1 \int\limits_{
1}^{e^{\frac{\pi}{2}}} \cos (2\ln x)dx = -2R$ (2)* Đặt : $\begin{cases}u = \cos (2\ln x) \\ dv = dx \end{cases} \Rightarrow \begin{cases}du = -\frac{2}{x} \sin (2\ln x)dx \\ v = x \end{cases}$$ R = (xcos (2\ln x))\left| \begin{array}{l}e^{\frac{\pi}{2}}\\1\end{array} \right. + 2 \int\limits_{
1}^{e^{\frac{\pi}{2}}} \sin (2\ln x)dx = -e^{\frac{\pi}{2} } - 1 + 2S$ (3)Thay (3) vào (2) ta có : $S = 2e^{\frac{\pi}{2} }+2-4S \Rightarrow S =\frac{2}{5}(e^{\frac{\pi}{2} } + 1)$ (4)Thay (4) vào (1) ta có : $T = -1 + \frac{2}{5}(e^{\frac{\pi}{2} } + 1) =\frac{1}{5} (2e^{\frac{\pi}{2} }-3).$