Đáp án:
a) \( \dfrac{{{{\sin }^3}x}}{3} + C\)
b) $ - \dfrac{2}{{\ln 5}}{\left( {\dfrac{1}{5}} \right)^x} + \dfrac{5}{{\ln 2}}{\left( {\dfrac{1}{2}} \right)^x} + C$
Giải thích các bước giải:
a) \(\int {{{\sin }^2}x\cos xdx} \)
Đặt \(\sin x = t \Rightarrow \cos xdx = dt\)
\( \Rightarrow I = \int {{t^2}dt} = \dfrac{{{t^3}}}{3} + C = \dfrac{{{{\sin }^3}x}}{3} + C\)
b)
\(\begin{array}{l}\int {\dfrac{{{2^{x + 1}} - {5^{x + 1}}}}{{{{10}^x}}}dx} = \int {\left( {\dfrac{{{{2.2}^x} - {{5.5}^x}}}{{{{10}^x}}}} \right)dx} \\ = \int {\left( {2.\dfrac{{{2^x}}}{{{{10}^x}}} - 5.\dfrac{{{5^x}}}{{{{10}^x}}}} \right)dx} = \int {\left[ {2.{{\left( {\dfrac{1}{5}} \right)}^x} - 5.{{\left( {\dfrac{1}{2}} \right)}^x}} \right]dx} \\ = 2\int {{{\left( {\dfrac{1}{5}} \right)}^x}dx} - 5\int {{{\left( {\dfrac{1}{2}} \right)}^x}dx} \\ = 2.\dfrac{{{{\left( {\dfrac{1}{5}} \right)}^x}}}{{\ln \dfrac{1}{5}}} - 5.\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^x}}}{{\ln \dfrac{1}{2}}} + C\\ = - \dfrac{2}{{\ln 5}}{\left( {\dfrac{1}{5}} \right)^x} + \dfrac{5}{{\ln 2}}{\left( {\dfrac{1}{2}} \right)^x} + C\end{array}\)
