Bài 24. Tính các tích phân sau :
a) \(\int\limits_1^2 {{x^2}{e^{{x^3}}}dx;} \) b) \(\int\limits_1^3 {{1 \over x}} {\left( {\ln x} \right)^2}dx;\)
c) \(\int\limits_0^{\sqrt 3 } {x\sqrt {1 + {x^2}} } dx;\)
\(d)\,\int\limits_0^1 {{x^2}{e^{3{x^3}}}dx;} \) \(e)\,\int\limits_0^{{\pi \over 2}} {{{\cos x} \over {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}}}} dx.\)
a) Đặt \(u = {x^3} \Rightarrow du = 3{x^2}dx \Rightarrow {x^2}dx = {{du} \over 3}\)
\(\int\limits_1^2 {{x^2}{e^{{x^3}}}dx = {1 \over 3}} \int\limits_1^8 {{e^u}du = \left. {{1 \over 3}{e^u}} \right|_1^8} = {1 \over 3}\left( {{e^8} - e} \right)\)
b) Đặt \(u = \ln x \Rightarrow du = {{dx} \over x}\)
Advertisements (Quảng cáo)
\(\int\limits_1^3 {{1 \over x}} {\left( {\ln x} \right)^2}dx = \int\limits_0^{\ln 3} {{u^2}du = \left. {{{{u^3}} \over 3}} \right|} _0^{\ln 3} = {1 \over 3}{\left( {\ln 3} \right)^3}\)
c) Đặt \(u = \sqrt {1 + {x^2}} \Rightarrow {u^2} = 1 + {x^2} \Rightarrow udu = xdx\)
\(\int\limits_0^{\sqrt 3 } {x\sqrt {1 + {x^2}} } dx = \int\limits_1^2 {u.udu = \left. {{{{u^3}} \over 3}} \right|} _1^2 = {7 \over 3}\)
d) Đặt \(u = 3{x^3} \Rightarrow du = 9{x^2}dx \Rightarrow {x^2}dx = {1 \over 9}du\)
\(\int\limits_0^1 {{x^2}{e^{3{x^3}}}dx = {1 \over 9}} \int\limits_0^3 {{e^u}du} = \left. {{1 \over 9}{e^u}} \right|_0^3 = {1 \over 9}\left( {{e^3} - 1} \right)\)
e) Đặt \(u = 1 + {\mathop{\rm s}\nolimits} {\rm{inx}} \Rightarrow du = \cos xdx\)
\(\int\limits_0^{{\pi \over 2}} {{{\cos xdx} \over {1 + {\mathop{\rm s}\nolimits} {\rm{inx}}}}} = \int\limits_1^2 {{{du} \over u}} = \left. {\ln \left| u \right|} \right|_1^2 = \ln 2\)