CÂU HỎI: Cho \( \smallint \sqrt[6]{{1 – {{\cos }^3}x}}\sin x{\cos ^5}xdx = 2\left( {\frac{{{t^\alpha }}}{\alpha } – \frac{{{t^\beta }}}{\beta }} \right)+C\) với \(t = \sqrt[6]{{1 – {{\cos }^3}x}}\)  . Tỉ số \( \frac{\alpha }{\beta }\)

Ta có:

\(
\smallint \sqrt[6]{{1 – {{\cos }^3}x}}\sin x{\cos ^5}xdx = \smallint \sqrt[6]{{1 – {{\cos }^3}x}}.{\cos ^3}x.\sin x.{\cos ^2}xdx\)

Đặt

\(\begin{array}{l}
t = \sqrt[6]{{1 – {{\cos }^3}x}}\\
{\cos ^3}x = 1 – {t^6}\\
\smallint t.(1 – {t^6}).2{t^5}.dt = \smallint (2{t^6} – 2{t^{12}}).dt = \frac{{2{t^7}}}{7} – {\frac{{2t}}{{13}}^{13}} + C = 2\left( {\frac{{{t^7}}}{7} – \frac{{{t^{13}}}}{{13}}} \right) + C\\
\Rightarrow \alpha = 7;\beta = 13 \Rightarrow \frac{\alpha }{\beta } = \frac{7}{{13}}
\end{array}\)