Bài 3.10: Tính các tích phân sau:
a) \(\int\limits_0^1 {({y^3} + 3{y^2} – 2)dy} \)
b)\(\int\limits_1^4 {(t + {1 \over {\sqrt t }}} – {1 \over {{t^2}}})dt\)
c) \(\int\limits_0^{{\pi \over 2}} {(2\cos x – \sin 2x)dx} \)
d) \(\int\limits_0^1 {{{({3^s} – {2^s})}^2}ds} \)
e) \(\int\limits_0^{{\pi \over 3}} {\cos 3xdx} + \int\limits_{{\pi \over 3}}^{{{3\pi } \over 2}} {\cos 3xdx} + \int\limits_{{{3\pi } \over 2}}^{{{5\pi } \over 2}} {\cos 3xdx} \)
g)\(\int\limits_0^3 {|{x^2} – x – 2|dx} \)
h) \(\int\limits_\pi ^{{{5\pi } \over 4}} {{{\sin x – \cos x} \over {\sqrt {1 + \sin 2x} }}} dx\)
i) \(\int\limits_0^4 {{{4x – 1} \over {\sqrt {2x + 1} + 2}}} dx\)
a) \( – {3 \over 4}\)
b) \({{35} \over 4}\)
c) 1
d) \({4 \over {\ln 3}} – {{10} \over {\ln 6}} + {3 \over {2\ln 2}}\)
e) \( – {1 \over 3}\)
g) \({{31} \over 6}\) .
HD: \(\int\limits_0^3 {|{x^2} – x – 2|dx }\)
\({= \int\limits_0^2 { – ({x^2} – x – 2)dx + \int\limits_2^3 {({x^2} – x – 2)dx} } } \)
h) \({1 \over 2}\ln 2\) .
HD: \(\int\limits_\pi ^{{{5\pi } \over 4}} {{{\sin x – \cos x} \over {\sqrt {1 + \sin 2x} }}} dx\)
\(= \int\limits_\pi ^{{{5\pi } \over 4}} {{{\sin x – \cos x} \over {|\sin x + \cos x|}}} dx = \int\limits_\pi ^{{{5\pi } \over 4}} {{{d(\sin x + \cos x)} \over {\sin x + \cos x}}} \)
i) \({{34} \over 3} + 10\ln {3 \over 5}\) .
HD: Đặt \(t = \sqrt {2x + 1} \)
Bài 3.11: Tính các tích phân sau bằng phương pháp đổi biến:
a) \(\int\limits_1^2 {x{{(1 – x)}^5}dx} \) (đặt t = 1 – x)
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b) \(\int\limits_0^{\ln 2} {\sqrt {{e^x} – 1} dx} \) (đặt \(t = \sqrt {{e^x} – 1} \))
c) \(\int\limits_1^9 {x\root 3 \of {1 – x} dx} \) (đặt \(t = \root 3 \of {1 – x} \))
d) \(\int\limits_{ – 1}^1 {{{2x + 1} \over {\sqrt {{x^2} + x + 1} }}} dx\) (đặt \(u = \sqrt {{x^2} + x + 1} \) )
e) \(\int\limits_1^2 {{{\sqrt {1 + {x^2}} } \over {{x^4}}}} dx\) (đặt \(t = {1 \over x}\))
g) \(\int\limits_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx} \) (đặt \(x = \pi – t\) )
h) \(\int\limits_{ – 1}^1 {{x^2}{{(1 – {x^3})}^4}dx} \)
i) \(\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} \) (đặt \(x = \tan u\) )
a) \( – {{13} \over {42}}\)
b) \(2 – {\pi \over 2}\)
c) \( – {{468} \over 7}\)
d) \(2(\sqrt 3 – 1)\)
e) \( – {1 \over 3}({{5\sqrt 5 } \over 8} – 2\sqrt 2 )\)
g) \({{{\pi ^2}} \over 4}\) .
HD: Đặt \(x = \pi – t\) , ta suy ra:
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\(\int\limits_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx} = {\pi \over 2}\int\limits_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx = {\pi \over 2}\int\limits_0^\pi {{{ – d(\cos x)} \over {1 + {{\cos }^2}x}}} \)
Vậy \(\int\limits_0^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx} = {\pi \over 2}\int\limits_{ – 1}^1 {{{dt} \over {1 + {t^2}}}} \) .
Đặt tiếp t = tan u
h) \({{{2^5}} \over {15}}\) .
HD: Đặt t = 1 – x3
i) \({\pi \over 4}\)
Bài 3.12: Áp dụng phương pháp tính tích phân từng phần, hãy tính các tích phân sau:
a) \(\int\limits_0^{{\pi \over 2}} {x\cos 2xdx} \)
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) \(\int\limits_{{1 \over 2}}^2 {(1 + x – {1 \over x}){e^{x + {1 \over x}}}dx} \)
g) \(\int\limits_0^{{\pi \over 2}} {x\cos x{{\sin }^2}xdx} \)
h) \(\int\limits_0^1 {{{x{e^x}} \over {{{(1 + x)}^2}}}} dx\)
i) \(\int\limits_1^e {{{1 + x\ln x} \over x}} {e^x}dx\)
a) \( – {1 \over 2}\)
b) \({1 \over 4}({3 \over 4} – {{\ln 2} \over 2})\)
c) \({3 \over 2}\ln 3 – 1\)
d) \(3\ln 3 – 6\ln 2\)
e) \({3 \over 2}{e^{{5 \over 2}}}\) .
HD: \(\int\limits_{{1 \over 2}}^2 {(1 + x – {1 \over x}){e^{x + {1 \over x}}}dx = } \int\limits_{{1 \over 2}}^2 {{e^{x + {1 \over x}}}} dx + \int\limits_{{1 \over 2}}^2 {(x – {1 \over x}){e^{x + {1 \over x}}}dx} \)
Tính tích phân từng phần: \(\int\limits_{{1 \over 2}}^2 {{e^{x + {1 \over x}}}dx = x{e^{x + {1 \over x}}}\left| {\matrix{2 \cr {{1 \over 2}} \cr} } \right.} – \int\limits_{{1 \over 2}}^2 {(x – {1 \over x}){e^{x + {1 \over x}}}dx} \)
g) \({\pi \over 6} – {2 \over 9}\)
HD: Đặt \(u = x,dv = \cos x{\sin ^2}xdx\)
h) \({e \over 2} – 1\). HD: \(\int\limits_0^1 {{{x{e^x}} \over {{{(1 + x)}^2}}}} dx = \int\limits_0^1 {{{{e^x}} \over {1 + x}}dx} – \int\limits_0^1 {{{{e^x}} \over {{{(1 + x)}^2}}}dx} \) và tính tích phân từng phần :
\(\int\limits_0^1 {{{x{e^x}} \over {{{(1 + x)}^2}}}} dx = {{ – {e^x}} \over {1 + x}}\left| {\matrix{
1 \cr 0 \cr} + } \right.\int\limits_0^1 {{{{e^x}} \over {1 + x}}dx} \)
i) ee . HD: Tương tự câu g)
Bài 3.13: Tính các tích phân sau đây:
a) \(\int\limits_0^{{\pi \over 2}} {(x + 1)\cos (x + {\pi \over 2}} )dx\)
b) \(\int\limits_0^1 {{{{x^2} + x + 1} \over {x + 1}}{{\log }_2}(x + 1)dx} \)
c) \(\int\limits_{{1 \over 2}}^1 {{{{x^2} – 1} \over {{x^4} + 1}}} dx\) (đặt \(t = x + {1 \over x}\))
d)\(\int\limits_0^{{\pi \over 2}} {{{\sin 2xdx} \over {3 + 4\sin x – \cos 2x}}} \)
a) – 2
b) \({1 \over {2\ln 2}}({1 \over 2} + {\ln ^2}2)\) . HD:\({{{x^2} + x + 1} \over {x + 1}}{\log _2}(x + 1) = {1 \over {\ln 2}}{\rm{[}}x\ln (x + 1) + {{\ln (x + 1)} \over {x + 1}}{\rm{]}}\)
c)\({1 \over {2\sqrt 2 }}\ln {{6 – \sqrt 2 } \over {6 + \sqrt 2 }}\) . HD: Đặt \(t = x + {1 \over x}\) , ta nhận được:
\(\int\limits_{{5 \over 2}}^2 {{{dt} \over {{t^2} – 2}} = {1 \over {2\sqrt 2 }}} \ln |{{t – \sqrt 2 } \over {t + \sqrt 2 }}|\left| {\matrix{2 \cr {{5 \over 2}} \cr} } \right. = {1 \over {2\sqrt 2 }}\ln {{6 – \sqrt 2 } \over {6 + \sqrt 2 }}\)
d) \(\ln 2 – {1 \over 2}\) . HD: \(\int\limits_0^{{\pi \over 2}} {{{\sin 2xdx} \over {3 + 4\sin x – \cos 2x}} = } \int\limits_0^{{\pi \over 2}} {\sin x.{{d(\sin x + 1)} \over {{{(\sin x + 1)}^2}}}} = \ln 2 – {1 \over 2}\)
