Bài tập 50 trang 176 SGK Toán 12 NC

a) Đặt \(\left\{ \begin{array}{l}
u = {x^2}\\
dv = \sin 2xdx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = 2xdx\\
v =  - \frac{1}{2}\cos 2x
\end{array} \right.\)

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\int_0^{\frac{\pi }{2}} {{x^2}\sin 2xdx} \\
 = \left. { - \frac{1}{2}{x^2}\cos 2x} \right|_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {{x^2}\cos 2xdx} 
\end{array}\\
{ = \frac{{{\pi ^2}}}{8} + \int_0^{\frac{\pi }{2}} {x\cos 2xdx\left( 1 \right)} }
\end{array}\)

Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = \cos 2xdx
\end{array} \right. \)

\(\Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = \frac{1}{2}\sin 2x
\end{array} \right.\)

Do đó: 

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\int_0^{\frac{\pi }{2}} {x\cos 2xxdx} \\
 = \left. {\frac{1}{2}x\sin 2x} \right|_0^{\frac{\pi }{2}} - \frac{1}{2}\int_0^{\frac{\pi }{2}} {\sin 2xdx} 
\end{array}\\
{ = \left. {\frac{1}{4}\cos 2x} \right|_0^{\frac{\pi }{2}} =  - \frac{1}{2}\left( 2 \right)}
\end{array}\)

Thay (2) vào (1) ta được:

\(\int_0^{\frac{\pi }{2}} {{x^2}\sin 2xdx}  = \frac{{{\pi ^2}}}{8} - \frac{1}{2}\)

b) Đặt \(u = 2{x^2} + 1\)

\(\Rightarrow du = 4xdx \Rightarrow xdx = \frac{{du}}{4}\)

\(\begin{array}{l}
\int_1^2 {x(2{x^2} + 1)dx} \\
 = \frac{1}{4}\int_3^9 {udu}  = \left. {\frac{1}{8}{u^2}} \right|_3^9 = 9
\end{array}\)

c) Đặt \(u = {x^2} - 2x \)

\(\Rightarrow du = 2(x - 1)dx \Rightarrow (x - 1)dx = \frac{{du}}{2}\)

\(\begin{array}{l}
\int_2^3 {(x - 1){e^{{x^2} - 2x}}dx} \\
 = \frac{1}{2}\int_0^3 {{e^u}du} \\
 = \left. {\frac{1}{2}{e^u}} \right|_0^3 = \frac{1}{2}({e^3} - 1)
\end{array}\)

-- Mod Toán 12 HỌC247