What is the solution for y^'=sqrt(tan(x))(sec(x))^4 ?

The solution for y^'=sqrt(tan(x))(sec(x))^4 is y= 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+c_{1}

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

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}

Solution steps

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

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

then

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}

n = 1 \sum \infty 7^{n + 1}

\int x^{3} e^{7 x} d x

\frac{d}{d x} (\frac{x}{1 + ( 1.5574 + 5.21 x ^{2} ) ^{2}})

x \to 0 + lim (e^{- \frac{1}{x}})

y^{^{'}} - 3 y = 6, y (0) = 1

What is the solution for y^'=sqrt(tan(x))(sec(x))^4 ?
The solution for y^'=sqrt(tan(x))(sec(x))^4 is y= 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+c_{1}