What is the solution for y^'=sin(x^2) ?

The solution for y^'=sin(x^2) is y=sqrt(pi/2)\S(sqrt(2/pi)x)+c_{1}

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y^'=sin(x^2)

y^{^{'}} = sin (x^{2})

y = \frac{π}{2} § (\frac{2}{π} x) + c_{1}

Solution steps

y^{'} (x) = sin (x^{2})

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

then

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

y = \int sin (x^{2}) d x

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x) + c_{1}

y = \frac{π}{2} § (\frac{2}{π} x) + c_{1}

n = 0 \sum \infty \frac{n + 4}{3 n + 2}

\int (z - 3)^{n} d z

\int_{- 1}^{1} (7 x^{3} - 2 x^{2} + 5 x - 4) d x

\frac{d y}{d x} = \frac{2 x + 1}{2 y}

f (x) = \frac{1}{x + 3}

What is the solution for y^'=sin(x^2) ?
The solution for y^'=sin(x^2) is y=sqrt(pi/2)\S(sqrt(2/pi)x)+c_{1}