Giải bài 3.18 trang 171 - SBT Đại số và Giải tích lớp 12

a) Đặt \(\left\{ \begin{aligned} & u=x \\ & dv=\cos 2xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=\frac{1}{2}\sin 2x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{aligned} & \int\limits_{0}^{\frac{\pi }{2}}{x\cos 2xdx}=\frac{1}{2}x\sin 2x\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right.-\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\sin 2xdx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0+\frac{1}{4}\cos 2x\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=+\frac{1}{4}\left( -1-1 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{1}{2} \\ \end{aligned} \)

b) Đặt \(\left\{ \begin{aligned} & u=x \\ & dv=e^{- 2x}dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\frac{1}{2}e^{- 2x} \\ \end{aligned} \right. \)

\(\begin{aligned} & \int\limits_{0}^{ln2}{xe^{-2x}dx}=-\frac{1}{2}xe^{-2x}\left| \begin{aligned} & ln2 \\ & 0 \\ \end{aligned} \right.+\frac{1}{2}\int\limits_{0}^{ln2}{e^{- 2x}dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{1}{2}\left( ln2.\dfrac{1}{4} -0 \right)-\frac{1}{4}e^{- 2x}\left| \begin{aligned} & ln2 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{1}{2}\left( ln2.\dfrac{1}{4} -0 \right)-\dfrac{1}{4}\left(\dfrac{1}{4}-1\right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{16}-\dfrac{1}{8}ln2 \\ \end{aligned} \)

c) Đặt \(\left\{ \begin{aligned} & u=\ln \left( 2x+1 \right) \\ & dv=dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\frac{2dx}{2x+1} \\ & v=x \\ \end{aligned} \right. \)

ta có: 

\(\begin{aligned} & \int\limits_{0}^{1}{\ln \left( 2x+1 \right)dx}=x\ln \left( 2x+1 \right)\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-\int\limits_{0}^{1}{\frac{2x}{2x+1}dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\ln 3-\int\limits_{0}^{1}{\left( 1-\frac{1}{2x+1} \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\ln 3-\left[ x-\frac{1}{2}\ln \left( 2x+1 \right) \right]\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\ln 3-\left( 1-\frac{1}{2}\ln 3 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{2}\ln 3-1 \\ \end{aligned} \)

d) Đặt \(\left\{ \begin{aligned} & u=\ln \left( x-1 \right) -\ln \left( x+1 \right)\\ & dv=dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx \\ & v=x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{aligned} & \int\limits_{2}^{3}{[\ln \left( x-1 \right) -\ln \left( x+1 \right)]dx}=[x\ln \left( x-1 \right)-xln(x+1)]\left| \begin{aligned} & 3 \\ & 2 \\ \end{aligned} \right.-\int\limits_{2}^{3}{\left(\frac{x}{x-1}-\frac{x}{x+1}\right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\ln 3-3 \ln2-\int\limits_{2}^{3}{\left( \frac{1}{x-1}+\frac{1}{x+1} \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\ln 3-3 \ln2-\left[ \ln \left( x-1 \right) +ln(x+1)\right]\left| \begin{aligned} & 3 \\ & 2 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\ln 3-3 \ln2-\left( 3\ln2 -\ln3 \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3\ln 3-6\ln 2 \\ \end{aligned} \)

e) \(I=\int\limits_{\frac{1}{2}}^{2}{\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\frac{1}{x}}}dx}=\int\limits_{\frac{1}{2}}^{2}{{{e}^{x+\frac{1}{x}}}dx}+\int\limits_{\frac{1}{2}}^{2}{\left( x-\dfrac{1}{x} \right){{e}^{x+\frac{1}{x}}}dx}={{I}_{1}}+{{I}_{2}}\)

Tính tích phân \(I_1\)

Đặt \(\left\{ \begin{aligned} & u={{e}^{x+\frac{1}{x}}} \\ & dv=dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\left( 1-\frac{1}{{{x}^{2}}} \right){e}^{x+\frac{1}{x}} \\ & v=x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{aligned} & {{I}_{1}}=x{{e}^{x+\frac{1}{x}}}\left| \begin{aligned} & 2 \\ & \frac{1}{2} \\ \end{aligned} \right.-\int\limits_{\frac{1}{2}}^{2}{x\left( 1-\frac{1}{{{x}^{2}}} \right){{e}^{x+\frac{1}{x}}}dx} \\ & \,\,\,\,\,\,=2{{e}^{\frac{5}{2}}}-\frac{1}{2}{{e}^{\frac{5}{2}}}-{{I}_{2}} \\ & \,\,\,\,\,=\frac{3}{2}{{e}^{\frac{5}{2}}}-{{I}_{2}} \\ \end{aligned} \)

Suy ra:

 \(I=\dfrac{3}{2}{{e}^{\frac{5}{2}}}\)

g) Đặt \(\left\{ \begin{aligned} & u=x \\ & dv=\cos x{{\sin }^{2}}xdx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=\frac{1}{3}{{\sin }^{3}}x \\ \end{aligned} \right. \)

Ta có: 

\(\begin{aligned} & \int\limits_{0}^{\frac{\pi }{2}}{x\cos x{{\sin }^{2}}xdx}=\frac{1}{3}x{{\sin }^{3}}x\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right.-\frac{1}{3}\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}xdx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\pi }{6}-\frac{1}{3}J \\ \end{aligned} \)

trong đó 

\(\begin{aligned} & J=\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}xdx}=\int\limits_{0}^{\frac{\pi }{2}}{\left( \frac{3}{4}\sin x-\frac{1}{4}\sin 3x \right)dx} \\ & \,\,\,=\left( -\frac{3}{4}\cos x+\frac{1}{12}\cos 3x \right)\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,=\frac{3}{4}-\frac{1}{12}=\frac{2}{3} \\ \end{aligned} \)

Vậy \(\int\limits_{0}^{\frac{\pi }{2}}{x\cos x{{\sin }^{2}}xdx}=\dfrac{\pi }{6}-\dfrac{2}{9}\).

Ghi nhớ: 

Phương pháp tích phân từng phần:

\(\int\limits_{a}^{b}{u\left( x \right)v'\left( x \right)dx}=\left[ u\left( x \right)v\left( x \right) \right]\left| \begin{align} & b \\ & a \\ \end{align} \right.-\int\limits_{a}^{b}{u'\left( x \right)v\left( x \right)dx} \)