Вычислить неопределенный интеграл (с объяснением)
[tex] \int\limits { \frac{sin^3xdx}{ \sqrt[3]{cox^2x} } } \,[/tex]
[tex] \int\limits { \frac{sin^3xdx}{ \sqrt[3]{cox^2x} } } \,[/tex]
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Замена cos x = y, dy = -sin x dx, тогда sin^2 x = 1 - cos^2 x = 1 - y^2корень кубический (cos^2 x) = (cos x)^(2/3) = y^(2/3)
Int (sin^2 x*sin x dx)/(cos x)^(2/3) =
= -Int (1 - y^2)/y^(2/3) dy = Int (y^2 - 1)*y^(-2/3) dy = Int y^(4/3) dy - Int y^(-2/3) dy =
= 3/7*y^(7/3) - 3y^(1/3) + C = 3/7*(cos x)^(7/3) - 3(cos x)^(1/3) + C