Câu 3.27: Tính các nguyên hàm sau:
a) \(\int {(2x – 3)\sqrt {x – 3} dx} \) , đặt \(u = \sqrt {x – 3} \)
b) \(\int {{x \over {{{(1 + {x^2})}^{{3 \over 2}}}}}} dx\) , đặt \(u = \sqrt {{x^2} + 1} \)
c) \(\int {{{{e^x}} \over {{e^x} + {e^{ – x}}}}} dx\) , đặt \(u = {e^{2x}} + 1\)
d) \(\int {{1 \over {\sin x – \sin a}}} dx\)
e) \(\int {\sqrt x \sin \sqrt x } dx\) , đặt \(t = \sqrt x \)
g)\(\int {x\ln {x \over {1 + x}}} dx\)
a) \({2 \over 5}{(x – 3)^{{3 \over 2}}}(2x – 1) + C\)
b)\( – {1 \over {\sqrt {1 + {x^2}} }} + C\)
c) \({1 \over 2}\ln ({e^{2x}} + 1) + C\)
d) \({1 \over {\cos a}}\ln |{{\sin {{x – a} \over 2}} \over {\cos {{x – a} \over 2}}}| + C\) . HD: Ta có:\(\cos a = \cos ({{x – a} \over 2} – {{x + a} \over 2})\)
e) \( – 2x\cos \sqrt x + 4\sqrt x \sin \sqrt x + 4\cos \sqrt x + C\)
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g) \({{{x^2}} \over 2}\ln {x \over {1 + x}} + {1 \over 2}\ln |1 + x| – {1 \over 2}x + C\)
Bài 3.28: Tính các tích phân sau:
a) \(\int\limits_0^1 {{{(y – 1)}^2}\sqrt y } dy\), đặt \(t = \sqrt y \)
b) \(\int\limits_1^2 {({z^2} + 1)\root 3 \of {{{(z – 1)}^2}} } dz\) , đặt \(u = \root 3 \of {{{(z – 1)}^2}} \)
c) \(\int\limits_1^e {{{\sqrt {4 + 5\ln x} } \over x}} dx\)
d) \(\int\limits_0^{{\pi \over 2}} {({{\cos }^5}\varphi } – {\sin ^5}\varphi )d\varphi \)
e) \(\int\limits_0^\pi {{{\cos }^3}\alpha \cos 3\alpha } d\alpha \)
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a) \({{16} \over {105}}\)
b) \(2{{49} \over {220}}\)
c) \({{38} \over {15}}\) .
HD: \(\int\limits_1^e {{{\sqrt {4 + 5\ln x} } \over x}} dx = {1 \over 5}\int\limits_1^e {{{(4 + 5\ln x)}^{{1 \over 2}}}d(4 + 5\ln x)} \)
d) 0
e)\({\pi \over 8}\) .
HD: Dùng công thức hạ bậc đối với \({\cos ^3}x\)
Câu 3.29: Tính các tích phân sau:
a) \(\int\limits_0^{{\pi \over 4}} {\cos 2x} .{\cos ^2}xdx\)
b) \(\int\limits_{{1 \over 2}}^1 {{{{e^x}} \over {{e^{2x}} – 1}}} dx\)
c) \(\int\limits_0^1 {{{x + 2} \over {{x^2} + 2x + 1}}} \ln (x + 1)dx\)
d) \(\int\limits_0^{{\pi \over 4}} {{{x\sin x + (x + 1)\cos x} \over {x\sin x + \cos x}}} dx\)
a) \({1 \over 4}(1 + {\pi \over 4})\) . HD: \({{1 + \cos 2x} \over 2} = {\cos ^2}x\)
b) \({1 \over 2}\ln {{(e – 1)(\sqrt e + 1)} \over {(e + 1)(\sqrt e – 1)}}\) . HD:\({{{e^x}} \over {{e^{2x}} – 1}} = {1 \over 2}({{{e^x}} \over {{e^x} – 1}} – {{{e^x}} \over {{e^x} + 1}})\)
c) \({1 \over 2}({\ln ^2}2 – \ln 2 + 1)\) . HD: \({{x + 2} \over {{x^2} + 2x + 1}}\ln (x + 1) = {{\ln (x + 1)} \over {x + 1}} + {{\ln (x + 1)} \over {{{(x + 1)}^2}}}\)
d) \({\pi \over 4} + \ln (1 + {\pi \over 4}) – {1 \over 2}\ln 2\) .
HD: \({{x\sin x + (x + 1)\cos x} \over {x\sin x + \cos x}} = 1 + {{x\cos x} \over {x\sin x + \cos x}}\) và \(d(x\sin x + \cos x) = x\cos xdx\)
