Bài tập 3.43 trang 180 SBT Toán 12
Bài tập 3.43 trang 180 SBT Toán 12
Tính các nguyên hàm sau:
a) \(\int (2x - 3)\sqrt {x - 3} dx\), đặt \(u = \sqrt {x - 3} \)
b) \(\int \frac{x}{{{{(1 + {x^2})}^{\frac{3}{2}}}}}dx\), đặt \(u = \sqrt {{x^2} + 1} \)
c) \(\int \frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}dx\), đặt \(u = {e^{2x}} + 1\)
d) \(\int \frac{1}{{\sin x - \sin a}}dx\)
a) Đặt \(u = \sqrt {x - 3} \)
\( \Rightarrow {u^2} = x - 3 \Rightarrow 2udu = dx\)
\(\begin{array}{l}
\Rightarrow \int (2x - 3)\sqrt {x - 3} dx\\
= \int \left[ {2\left( {{u^2} + 3} \right) - 3} \right].u.2udu\\
= 2\int {u^2}\left( {2{u^2} + 3} \right)du\\
= 2\int \left( {2{u^4} + 3{u^2}} \right)du\\
= 2\left( {2.\frac{{{u^5}}}{5} + 3.\frac{{{u^3}}}{3}} \right) + C\\
= \frac{4}{5}{u^5} + {u^3} + C\\
= \frac{4}{5}.{\left( {\sqrt {x - 3} } \right)^5} + {\left( {\sqrt {x - 3} } \right)^3} + C\\
= \frac{4}{5}{\left( {x - 3} \right)^{\frac{5}{2}}} + {\left( {x - 3} \right)^{\frac{3}{2}}} + C
\end{array}\)
b) Đặt \(u = \sqrt {{x^2} + 1} \)
\( \Rightarrow {u^2} = {x^2} + 1 \Rightarrow udu = xdx\)
\(\begin{array}{l}
\Rightarrow \int \frac{x}{{{{(1 + {x^2})}^{\frac{3}{2}}}}}dx = \int \frac{{udu}}{{{u^3}}} = \int \frac{{du}}{{{u^2}}}\\
= - \frac{1}{u} + C = - \frac{1}{{\sqrt {1 + {x^2}} }} + C
\end{array}\)
c) Ta có:
\(\begin{array}{l}
\int \frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}dx = \int \frac{{{e^x}.{e^x}}}{{\left( {{e^x} + {e^{ - x}}} \right).{e^x}}}dx\\
= \int \frac{{{e^{2x}}}}{{{e^{2x}} + 1}}dx
\end{array}\)
Đặt \(u = {e^2}x + 1 \Rightarrow du = 2{e^{2x}}dx\)
Khi đó:
\(\begin{array}{l}
\int \frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}dx = \int \frac{{du}}{{2u}} = \frac{1}{2}\ln u\\
= \frac{1}{2}\ln \left( {{e^{2x}} + 1} \right) + C
\end{array}\)
d) Ta có:
\(\begin{array}{l}
\frac{1}{{\sin x - \sin a}} = \frac{1}{{2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\\
= \frac{{\cos a}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\\
= \frac{{\cos \left( {\frac{{x + a}}{2} - \frac{{x - a}}{2}} \right)}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\\
= \frac{{\cos \frac{{x + a}}{2}\cos \frac{{x - a}}{2} + \sin \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\\
= \frac{1}{{2\cos a}}\left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)
\end{array}\)
\(\begin{array}{l}
\Rightarrow \int \frac{1}{{\sin x - \sin a}}dx\\
= \frac{1}{{2\cos a}}\int \left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)dx
\end{array}\)
+) Tính \(J = \int \frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}}dx\)
\( = \int \ \frac{{2d\left( {\sin \frac{{x - a}}{2}} \right)}}{{\sin \frac{{x - a}}{2}}} = 2\ln \left| {\sin \frac{{x - a}}{2}} \right| + D\)
+) Tính \(K = \int \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}dx\)
\( = \int \frac{{ - 2d\left( {\cos \frac{{x + a}}{2}} \right)}}{{\cos \frac{{x + a}}{2}}} = - 2\ln \left| {\cos \frac{{x + a}}{2}} \right| + D\)
\(\begin{array}{l}
\Rightarrow I = \frac{1}{{2\cos a}}\left( {J + K} \right)\\
= \frac{1}{{2\cos a}}\left( {2\ln \left| {\sin \frac{{x - a}}{2}} \right| - 2\ln \left| {\cos \frac{{x + a}}{2}} \right|} \right) + C\\
= \frac{1}{{\cos a}}\ln \left| {\frac{{\sin \frac{{x - a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right| + C
\end{array}\)
-- Mod Toán 12
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