Bài 25 Trang 162 SGK Đại số và Giải tích 12 Nâng cao

Bài 25. Tính các tích phân sau :

a) \(\int\limits_0^{{\pi  \over 4}} {x\cos 2xdx;} \)         b) \(\int\limits_0^1 {{{\ln \left( {2 – x} \right)} \over {2 – x}}} dx;\)       

c) \(\int\limits_0^{{\pi  \over 2}} {{x^2}\cos xdx;} \)

\(d)\,\int\limits_0^1 {{x^2}\sqrt {{x^3} + 1} dx;} \)        \(e)\,\int\limits_1^e {{x^2}\ln xdx.} \)   

Giải

a) Đặt 

\(\left\{ \matrix{
u = x \hfill \cr
dv = \cos 2xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = {1 \over 2}\sin 2x \hfill \cr} \right.\)

Do đó \(\int\limits_0^{{\pi  \over 4}} {x\cos 2xdx = \left. {{1 \over 2}x\sin 2x} \right|_0^{{\pi  \over 4}}}  – {1 \over 2}\int\limits_0^{{\pi  \over 4}} {\sin 2xdx} \) 

\( = {\pi  \over 8} + \left. {{1 \over 4}\cos 2x} \right|_0^{{\pi  \over 4}} = {\pi  \over 8} + {1 \over 4}\left( { – 1} \right) = {\pi  \over 8} – {1 \over 4}.\)                                 

b) Đặt \(u = \ln \left( {2 – x} \right) \Rightarrow du = {{ – 1} \over {2 – x}}dx\)

\(\int\limits_0^1 {{{\ln \left( {2 – x} \right)} \over {2 – x}}} dx =  – \int\limits_{\ln 2}^0 {udu}  = \int\limits_0^{\ln 2} {udu}  = \left. {{{{u^2}} \over 2}} \right|_0^{\ln 2} = {1 \over 2}{\left( {\ln 2} \right)^2}\)

c) Đặt 

\(\left\{ \matrix{
u = {x^2} \hfill \cr
dv = \cos xdx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = 2xdx \hfill \cr
v = {\mathop{\rm s}\nolimits} {\rm{inx}} \hfill \cr} \right.\)

Do đó \(I = \int\limits_0^{{\pi  \over 2}} {{x^2}\cos xdx = {x^2}} \left. {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^{{\pi  \over 2}} – 2\int\limits_0^{{\pi  \over 2}} {x\sin xdx = {{{\pi ^2}} \over 4}}  – 2{I_1}\)

Với \({I_1} = \int\limits_0^{{\pi  \over 2}} {x\sin xdx} \)

Đặt 

\(\left\{ \matrix{
u = x \hfill \cr
dv = \sin {\rm{x}}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = dx \hfill \cr
v = – \cos x \hfill \cr} \right.\)

Do đó \({I_1} =  – x\left. {\cos x} \right|_0^{{\pi  \over 2}} + \int\limits_0^{{\pi  \over 2}} {\cos xdx = \left. {{\mathop{\rm s}\nolimits} {\rm{inx}}} \right|_0^{{\pi  \over 2}}}  = 1\)

Vậy \(I = {{{\pi ^2}} \over 4} – 2\)

d) Đặt \(u = \sqrt {{x^3} + 1}  \Rightarrow {u^2} = {x^3} + 1 \Rightarrow 2udu = 3{x^2}dx \Rightarrow {x^2}dx = {2 \over 3}udu\)

\(\int\limits_0^1 {{x^2}\sqrt {{x^3} + 1} dx}  = {2 \over 3}\int\limits_1^{\sqrt 2 } {{u^2}du = \left. {{{2{u^3}} \over 9}} \right|} _1^{\sqrt 2 } = {2 \over 9}\left( {2\sqrt 2  – 1} \right)\)

e) Đặt 

\(\left\{ \matrix{
u = \ln x \hfill \cr
dv = {x^2}dx \hfill \cr} \right. \Rightarrow \left\{ \matrix{
du = {{dx} \over x} \hfill \cr
v = {{{x^3}} \over 3} \hfill \cr} \right.\)

Do đó \(\int\limits_1^e {{x^2}\ln xdx = \left. {{{{x^3}} \over 3}\ln x} \right|} _1^e – {1 \over 3}\int\limits_1^e {{x^2}dx = {{{e^3}} \over 3} – \left. {{1 \over 9}{x^3}} \right|} _1^e = {{2{e^3} + 1} \over 9}\)