Bài 3.18 trang 171 SBT giải tích 12

Áp dụng phương pháp tính tích phân từng phần, hãy tính các tích phân sau:

a) \(\int\limits_0^{\dfrac{\pi }{2}} {x\cos 2xdx} \)

b) \(\int\limits_0^{\ln 2} {x{e^{ - 2x}}dx} \)

c) \(\int\limits_0^1 {\ln (2x + 1)dx} \)

d) \(\int\limits_2^3 {{\rm{[}}\ln (x - 1) - \ln (x + 1){\rm{]}}dx} \)

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

g) \(\int\limits_0^{\dfrac{\pi }{2}} {x\cos x{{\sin }^2}xdx} \)

a) \(I = \int\limits_0^{\dfrac{\pi }{2}} {x\cos 2xdx} \)

Đặt \(\left\{ \begin{array}{l}u = x\\dv = \cos 2xdx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = dx\\v = \dfrac{{\sin 2x}}{2}\end{array} \right.\)

\( \Rightarrow I = \left. {\dfrac{{x\sin 2x}}{2}} \right|_0^{\dfrac{\pi }{2}} - \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\sin 2xdx} \) \( = \dfrac{1}{2}.\left. {\dfrac{{\cos 2x}}{2}} \right|_0^{\dfrac{\pi }{2}} =  - \dfrac{1}{4} - \dfrac{1}{4} =  - \dfrac{1}{2}\)

b) \(I = \int\limits_0^{\ln 2} {x{e^{ - 2x}}dx} \)

Đặt \(\left\{ \begin{array}{l}u = x\\dv = {e^{ - 2x}}dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = dx\\v =  - \dfrac{{{e^{ - 2x}}}}{2}\end{array} \right.\)

\( \Rightarrow I = \left. { - \dfrac{{x{e^{ - 2x}}}}{2}} \right|_0^{\ln 2} + \dfrac{1}{2}\int\limits_0^{\ln 2} {{e^{ - 2x}}dx} \) \( =  - \dfrac{{\ln 2.{e^{ - 2\ln 2}}}}{2} - \dfrac{1}{2}.\left. {\dfrac{{{e^{ - 2x}}}}{2}} \right|_0^{\ln 2}\) \( =  - \dfrac{{\ln 2}}{8} - \dfrac{3}{{16}}\)

c) \(I = \int\limits_0^1 {\ln (2x + 1)dx} \)

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {2x + 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{2}{{2x + 1}}dx\\v = x\end{array} \right.\)

\( \Rightarrow I = \left. {x\ln \left( {2x + 1} \right)} \right|_0^1 - \int\limits_0^1 {\dfrac{{2x}}{{2x + 1}}dx} \) \( = \ln 3 - \int\limits_0^1 {\left( {1 - \dfrac{1}{{2x + 1}}} \right)dx} \) \( = \ln 3 - \left. {\left( {x - \dfrac{{\ln \left( {2x + 1} \right)}}{2}} \right)} \right|_0^1\) \( = \ln 3 - \left( {1 - \dfrac{{\ln 3}}{2}} \right) = \dfrac{3}{2}\ln 3 - 1\)

d) \(I = \int\limits_2^3 {\left[ {\ln \left( {x - 1} \right) - \ln \left( {x + 1} \right)} \right]dx} \) \( = \int\limits_2^3 {\ln \left( {x - 1} \right)dx}  - \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \) \( = J - K\) với \(J = \int\limits_2^3 {\ln \left( {x - 1} \right)dx} \) và \(K = \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \).

+) Tính \(J = \int\limits_2^3 {\ln \left( {x - 1} \right)dx} \).

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x - 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{dx}}{{x - 1}}\\v = x\end{array} \right.\)

\( \Rightarrow J = \left. {x\ln \left( {x - 1} \right)} \right|_2^3 - \int\limits_2^3 {\dfrac{x}{{x - 1}}dx} \) \( = 3\ln 2 - \int\limits_2^3 {\left( {1 + \dfrac{1}{{x - 1}}} \right)dx} \) \( = 3\ln 2 - \left. {\left( {x + \ln \left( {x - 1} \right)} \right)} \right|_2^3\) \( = 3\ln 2 - 3 - \ln 2 + 2\) \( = 2\ln 2 - 1\).

+) Tính \(K = \int\limits_2^3 {\ln \left( {x + 1} \right)dx} \).

Đặt \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\dv = dx\end{array} \right.\) \( \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{dx}}{{x + 1}}\\v = x\end{array} \right.\)

\( \Rightarrow K = \left. {x\ln \left( {x + 1} \right)} \right|_2^3 - \int\limits_2^3 {\dfrac{x}{{x + 1}}dx} \) \( = 3\ln 4 - 2\ln 3 - \int\limits_2^3 {\left( {1 - \dfrac{1}{{x + 1}}} \right)dx} \) \( = 6\ln 2 - 2\ln 3 - \left. {\left( {x - \ln \left( {x + 1} \right)} \right)} \right|_2^3\) \( = 6\ln 2 - 2\ln 3 - 3 + \ln 4 + 2 - \ln 3\) \( = 8\ln 2 - 3\ln 3 - 1\).

\( \Rightarrow I = J - K\) \( = 2\ln 2 - 1 - \left( {8\ln 2 - 3\ln 3 - 1} \right)\) \( = 3\ln 3 - 6\ln 2\)

e) \(I = \int\limits_{\dfrac{1}{2}}^2 {\left( {1 + x - \dfrac{1}{x}} \right){e^{x + \dfrac{1}{x}}}dx} \)\( = \int\limits_{\dfrac{1}{2}}^2 {{e^{x + \dfrac{1}{x}}}} dx + \int\limits_{\dfrac{1}{2}}^2 {\left( {x - \dfrac{1}{x}} \right){e^{x + \dfrac{1}{x}}}dx} \) \( = J + K\) với \(J = \int\limits_{\dfrac{1}{2}}^2 {{e^{x + \dfrac{1}{x}}}} dx\) và \(K = \int\limits_{\dfrac{1}{2}}^2 {\left( {x - \dfrac{1}{x}} \right){e^{x + \dfrac{1}{x}}}dx} \)

+) Tính \(J = \int\limits_{\dfrac{1}{2}}^2 {{e^{x + \dfrac{1}{x}}}} dx\)

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

\( \Rightarrow J = \left. {x{e^{x + \dfrac{1}{x}}}} \right|_{\dfrac{1}{2}}^2 - \int\limits_{\dfrac{1}{2}}^2 {\left( {x - \dfrac{1}{x}} \right){e^{x + \dfrac{1}{x}}}dx} \) \( = \left. {x{e^{x + \dfrac{1}{x}}}} \right|_{\dfrac{1}{2}}^2 - K\) \( = 2{e^{\dfrac{5}{2}}} - \dfrac{1}{2}{e^{\dfrac{5}{2}}} - K = \dfrac{3}{2}{e^{\dfrac{5}{2}}} - K\)

Suy ra \(I = J + K\) \( = \dfrac{3}{2}{e^{\dfrac{5}{2}}} - K + K = \dfrac{3}{2}{e^{\dfrac{5}{2}}}\).

g) \(I = \int\limits_0^{\dfrac{\pi }{2}} {x\cos x{{\sin }^2}xdx} \)

Đặt  \(u = x,dv = \cos x{\sin ^2}xdx\) \( \Rightarrow du = dx\). Ta tìm \(v = \int {\cos x{{\sin }^2}xdx} \).

Đặt \(\sin x = t \Rightarrow dt = \cos xdx\)

\( \Rightarrow \int {\cos x{{\sin }^2}xdx}  = \int {{t^2}dt} \) \( = \dfrac{{{t^3}}}{3} + C = \dfrac{{{{\sin }^3}x}}{3} + C\)

Chọn \(v = \dfrac{{{{\sin }^3}x}}{3}\) ta có:

\(I = \int\limits_0^{\dfrac{\pi }{2}} {x\cos x{{\sin }^2}xdx} \)\( = \left. {\dfrac{{x{{\sin }^3}x}}{3}} \right|_0^{\dfrac{\pi }{2}} - \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^3}x}}{3}dx} \) \( = \dfrac{\pi }{6} - \dfrac{1}{3}\int\limits_0^{\dfrac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right)\sin xdx} \) \( = \dfrac{\pi }{6} - \dfrac{1}{3}J\)

Đặt \(\cos x = t \Rightarrow dt =  - \sin xdx\)

\( \Rightarrow J = \int\limits_1^0 {\left( {1 - {t^2}} \right).\left( { - dt} \right)} \) \( = \int\limits_0^1 {\left( {1 - {t^2}} \right)dt} \) \( = \left. {\left( {t - \dfrac{{{t^3}}}{3}} \right)} \right|_0^1 = \dfrac{2}{3}\)

Vậy \(I = \dfrac{\pi }{6} - \dfrac{1}{3}J\) \( = \dfrac{\pi }{6} - \dfrac{1}{3}.\dfrac{2}{3} = \dfrac{\pi }{6} - \dfrac{2}{9}\).