a) \(\displaystyle \int {(2x - 3)\sqrt {x - 3} dx} \), đặt \(\displaystyle u = \sqrt {x - 3} \)
Đặt \(\displaystyle u = \sqrt {x - 3} \)\(\displaystyle \Rightarrow {u^2} = x - 3 \Rightarrow 2udu = dx\)
\(\displaystyle \Rightarrow \int {(2x - 3)\sqrt {x - 3} dx} \) \(\displaystyle = \int {\left[ {2\left( {{u^2} + 3} \right) - 3} \right].u.2udu} \) \(\displaystyle = 2\int {{u^2}\left( {2{u^2} + 3} \right)du} \) \(\displaystyle = 2\int {\left( {2{u^4} + 3{u^2}} \right)du} \)
\(\displaystyle = 2\left( {2.\frac{{{u^5}}}{5} + 3.\frac{{{u^3}}}{3}} \right) + C\) \(\displaystyle = \frac{4}{5}{u^5} + {u^3} + C\) \(\displaystyle = \frac{4}{5}.{\left( {\sqrt {x - 3} } \right)^5} + {\left( {\sqrt {x - 3} } \right)^3} + C\) \(\displaystyle = \frac{4}{5}{\left( {x - 3} \right)^{\frac{5}{2}}} + {\left( {x - 3} \right)^{\frac{3}{2}}} + C\)
b) \(\displaystyle \int {\frac{x}{{{{(1 + {x^2})}^{\frac{3}{2}}}}}} dx\), đặt \(\displaystyle u = \sqrt {{x^2} + 1} \)
Đặt \(\displaystyle u = \sqrt {{x^2} + 1} \)\(\displaystyle \Rightarrow {u^2} = {x^2} + 1 \Rightarrow udu = xdx\)
\(\displaystyle \Rightarrow \int {\frac{x}{{{{(1 + {x^2})}^{\frac{3}{2}}}}}} dx\) \(\displaystyle = \int {\frac{{udu}}{{{u^3}}}} = \int {\frac{{du}}{{{u^2}}}} \) \(\displaystyle = - \frac{1}{u} + C = - \frac{1}{{\sqrt {1 + {x^2}} }} + C\)
c) \(\displaystyle \int {\frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}} dx\), đặt \(\displaystyle u = {e^{2x}} + 1\)
Ta có: \(\displaystyle \int {\frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}} dx\)\(\displaystyle = \int {\frac{{{e^x}.{e^x}}}{{\left( {{e^x} + {e^{ - x}}} \right).{e^x}}}dx} \) \(\displaystyle = \int {\frac{{{e^{2x}}}}{{{e^{2x}} + 1}}dx} \)
Đặt \(\displaystyle u = {e^{2x}} + 1 \Rightarrow du = 2{e^{2x}}dx\)
Khi đó \(\displaystyle \int {\frac{{{e^x}}}{{{e^x} + {e^{ - x}}}}} dx\) \(\displaystyle = \int {\frac{{du}}{{2u}}} = \frac{1}{2}\ln u\) \(\displaystyle = \frac{1}{2}\ln \left( {{e^{2x}} + 1} \right) + C\)
d) \(\displaystyle \int {\frac{1}{{\sin x - \sin a}}} dx\)
Ta có: \(\displaystyle \frac{1}{{\sin x - \sin a}}\)\(\displaystyle = \frac{1}{{2\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\) \(\displaystyle = \frac{{\cos a}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\)
\(\displaystyle = \frac{{\cos \left( {\frac{{x + a}}{2} - \frac{{x - a}}{2}} \right)}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\) \(\displaystyle = \frac{{\cos \frac{{x + a}}{2}\cos \frac{{x - a}}{2} + \sin \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}{{2\cos a\cos \frac{{x + a}}{2}\sin \frac{{x - a}}{2}}}\)
\(\displaystyle = \frac{1}{{2\cos a}}\left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)\)
\(\displaystyle \Rightarrow \int {\frac{1}{{\sin x - \sin a}}} dx\) \(\displaystyle = \frac{1}{{2\cos a}}\int {\left( {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}} + \frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right)dx} \)
+) Tính \(\displaystyle J = \int {\frac{{\cos \frac{{x - a}}{2}}}{{\sin \frac{{x - a}}{2}}}dx} \) \(\displaystyle = \int {\frac{{2d\left( {\sin \frac{{x - a}}{2}} \right)}}{{\sin \frac{{x - a}}{2}}}} \) \(\displaystyle = 2\ln \left| {\sin \frac{{x - a}}{2}} \right| + D\)
+) Tính \(\displaystyle K = \int {\frac{{\sin \frac{{x + a}}{2}}}{{\cos \frac{{x + a}}{2}}}dx} \) \(\displaystyle = \int {\frac{{ - 2d\left( {\cos \frac{{x + a}}{2}} \right)}}{{\cos \frac{{x + a}}{2}}}} \) \(\displaystyle = - 2\ln \left| {\cos \frac{{x + a}}{2}} \right| + D\)
\(\displaystyle \Rightarrow I = \frac{1}{{2\cos a}}\left( {J + K} \right)\) \(\displaystyle = \frac{1}{{2\cos a}}\left( {2\ln \left| {\sin \frac{{x - a}}{2}} \right| - 2\ln \left| {\cos \frac{{x + a}}{2}} \right|} \right) + C\) \(\displaystyle = \frac{1}{{\cos a}}\ln \left| {\frac{{\sin \frac{{x - a}}{2}}}{{\cos \frac{{x + a}}{2}}}} \right| + C\)
