Giải bài 11 trang 147 – SGK môn Giải tích lớp 12
Tính các tích phân sau bằng phương pháp tính tích phân từng phần:
a) \(\int\limits_{1}^{{{e}^{4}}}{\sqrt{x}\ln xdx}\);
b) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\dfrac{xdx}{{{\sin }^{2}}x}}\);
c) \(\int\limits_{0}^{\pi }{\left( \pi -x \right)\sin xdx}\);
d) \(\int\limits_{-1}^{0}{\left( 2x+3 \right){{e}^{-x}}dx}\).
Gợi ý:
Sử dụng tích phân từng phần.
a) \(\int\limits_{1}^{{{e}^{4}}}{\sqrt{x}\ln xdx}\);
Đặt \(\left\{ \begin{aligned} & \ln x=u \\ & \sqrt{x}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=\dfrac{dx}{x} \\ & v=\dfrac{2}{3}{{x}^{\frac{3}{2}}} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{1}^{{{e}^{4}}}{\sqrt{x}\ln xdx}&=\dfrac{2}{3}{{x}^{\frac{3}{2}}}.\ln x\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} {{e}^{4}} \\ \end{smallmatrix}} \right.-\dfrac{2}{3}\int\limits_{1}^{{{e}^{4}}}{\sqrt{x}dx} \\ & =\dfrac{2{{e}^{6}}}{3}.4-\dfrac{4}{9}{{x}^{\frac{3}{2}}}\left| _{\begin{smallmatrix} \\ 1 \end{smallmatrix}}^{\begin{smallmatrix} {{e}^{4}} \\ \end{smallmatrix}} \right. \\ & =\dfrac{8{{e}^{6}}}{3}-\dfrac{4{{e}^{6}}}{9}+\dfrac{4}{9} \\ & =\dfrac{20{{e}^{6}}}{9}+\dfrac{4}{9} \\ & \\ \end{aligned} \)
b) \(\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\dfrac{xdx}{{{\sin }^{2}}x}}\);
Đặt \(\left\{ \begin{aligned} & x=u \\ & \dfrac{dx}{{{\sin }^{2}}x}=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & dx=du \\ & v=-\cot x \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\dfrac{xdx}{{{\sin }^{2}}x}}&=-x\cot x\left| _{\frac{\pi }{6}}^{\frac{\pi }{2}} \right.+\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cot xdx} \\ & =\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx} \\ & =\dfrac{\pi \sqrt{3}}{6}+\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\dfrac{d\left( \sin x \right)}{\sin x}} \\ & =\dfrac{\pi \sqrt{3}}{6}+\ln \left| {\sin x} \right|\left| _{\frac{\pi }{6}}^{\frac{\pi }{2}} \right. \\ & =\dfrac{\pi \sqrt{3}}{6}+\ln 2 \\ \end{aligned} \)
c) \(\int\limits_{0}^{\pi }{\left( \pi -x \right)\sin xdx}\);
Đặt \(\left\{ \begin{aligned} & \pi -x=u \\ & \sin x dx=dv \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & du=-dx \\ & x=-\cos x \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{0}^{\pi }{\left( \pi -x \right)\sin xdx}&=\left( x-\pi \right)\cos x\left| _{\begin{smallmatrix} \\ 0 \end{smallmatrix}}^{\begin{smallmatrix} \pi \\ \end{smallmatrix}} \right.-\int\limits_{0}^{\pi }{\cos xdx} \\ & =\pi -\sin x\left| _{0}^{\pi } \right. \\ & =\pi \\ \end{aligned} \)
d) \(\int\limits_{-1}^{0}{\left( 2x+3 \right){{e}^{-x}}dx}\).
Đặt \(\left\{ \begin{aligned} & 2x+3=u \\ & {{e}^{-x}}dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & 2dx=du \\ & v=-{{e}^{-x}} \\ \end{aligned} \right. \)
\(\begin{aligned} \int\limits_{-1}^{0}{\left( 2x+3 \right){{e}^{-x}}dx}&=-\left( 2x+3 \right){{e}^{-x}}\left| _{-1}^{0} \right.+2\int\limits_{-1}^{0}{{{e}^{-x}}dx} \\ & =-3+e-2{{e}^{-x}}\left| _{-1}^{0} \right. \\ & =-3+e-2+2e \\ & =3e-5 \\ \end{aligned} \)