Đáp án:
$\begin{array}{l}
+ )\\
I = \int {{x^3}{{\left( {2 - 3{x^2}} \right)}^{2019}} = dx} \\
Đặt:2 - 3{x^2} = t\\
\Rightarrow \left\{ \begin{array}{l}
- 6xdx = dt \Rightarrow xdx = \frac{{ - dt}}{6}\\
{x^2} = \frac{{2 - t}}{3}
\end{array} \right.\\
\Rightarrow I = \int {{x^2}{{\left( {2 - 3{x^2}} \right)}^{2019}}.xdx} \\
= \int {\frac{{2 - t}}{3}.{t^{2019}}.} \frac{{ - dt}}{6}\\
= \int {\frac{1}{{18}}{t^{2020}} - \frac{1}{9}{t^{2019}}dt} \\
= \frac{{{t^{2021}}}}{{2021.18}} - \frac{{{t^{2020}}}}{{9.2020}} + c\\
= \frac{{{{\left( {2 - 3{x^2}} \right)}^{2021}}}}{{2021.18}} - \frac{{{{\left( {2 - 3{x^2}} \right)}^{2020}}}}{{9.2020}} + c\\
I = \int {x.\sqrt {1 - x} dx} \\
\sqrt {1 - x} = t\\
\Rightarrow \left\{ \begin{array}{l}
1 - x = {t^2} \Rightarrow x = 1 - {t^2}\\
- dx = 2tdt \Rightarrow dx = - 2tdt
\end{array} \right.\\
\Rightarrow I = \int {\left( {1 - {t^2}} \right).t.\left( { - 2t} \right)dt} \\
= \int {2{t^4} - 2{t^2}dt} \\
= \frac{{2{t^5}}}{5} - \frac{{2{t^3}}}{3} + c\\
= \frac{{2{{\left( {\sqrt {1 - x} } \right)}^5}}}{5} - \frac{{2{{\left( {\sqrt {1 - x} } \right)}^3}}}{3} + c
\end{array}$
ý 3 tương tự đặt √1-x=t
