y^'=sqrt(tan(x))(sec(x))^4

解

y^{^{'}} = tan (x) (sec (x))^{4}

y = \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + c_{1}

解答ステップ

y^{'} (x) = tan (x) (sec (x))^{4}

f^{^{'}} (x) = g (x)

であれば

f (x) = \int g (x) d x

y = \int tan (x) (sec (x))^{4} d x

\int tan (x) (sec (x))^{4} d x = \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + c_{1}

y = \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + c_{1}