Bài tập 9 trang 146 SGK Toán 12 NC

Bài tập 9 trang 146 SGK Toán 12 NC

a) Đặt \(\left\{ \begin{array}{l}
u = {x^2}\\
dv = \cos 2xdx\)

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

Do đó \(\int  {x^2}\cos 2xdx \)

\(= \frac{1}{2}{x^2}\sin 2x - \int  x\sin 2xdx \left( 1 \right)\)

Tính \(\int  x\sin 2xdx\)

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

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = dx}\\
{v =  - \frac{1}{2}\cos 2x}
\end{array}} \right.\)

\(\begin{array}{l}
 \Rightarrow \int x \sin 2xdx\\
 =  - \frac{1}{2}x\cos 2x + \frac{1}{2}\int {\cos } 2xdx\\
 =  - \frac{1}{2}x\cos 2x - \frac{1}{4}\sin 2x + C
\end{array}\)

Thay vào (1) ta được:

\(\begin{array}{l}
\int {{x^2}} \cos 2xdx\\
 = \frac{1}{2}{x^2}\sin 2x + \frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x + C
\end{array}\)

b) Đặt

\(\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{u = \ln x}\\
{dv = \sqrt x dx}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = \frac{{dx}}{x}}\\
{v = \frac{2}{3}{x^{\frac{3}{2}}}}
\end{array}} \right.}\\
{ \Rightarrow \int {\sqrt x } \ln xdx = \frac{2}{3}{x^{\frac{3}{2}}}\ln x - \frac{2}{3}\int {{x^{\frac{1}{2}}}} dx}\\
\begin{array}{l}
 = \frac{2}{3}{x^{\frac{3}{2}}}\ln x - \frac{2}{3}.\frac{2}{3}{x^{\frac{3}{2}}} + C\\
 = \frac{2}{3}\sqrt {{x^3}} \ln x - \frac{4}{9}\sqrt {{x^3}}  + C
\end{array}
\end{array}\)

c) Đặt \(u = sinx \Rightarrow du = cosxdx\)

\(\begin{array}{l}
 \Rightarrow \int {{{\sin }^4}} x\cos xdx = \int {{u^4}} du\\
 = \frac{{{u^5}}}{5} + C = \frac{1}{5}{\sin ^5}x + C.
\end{array}\)

d) Đặt \(u = {x^2} \Rightarrow du = 2xdx \)

\(\Rightarrow xdx = \frac{1}{2}du\)

\(\begin{array}{l}
 \Rightarrow \int x \cos \left( {{x^2}} \right)dx = \frac{1}{2}\int {\cos } udu\\
 = \frac{1}{2}\sin u + C = \frac{1}{2}{\rm{sin}}{{\rm{x}}^2} + C
\end{array}\)

-- Mod Toán 12