Calculus of a Single Variable, Chapter 6, 6.1, Section 6.1, Problem 46

To solve this differential equation, rewrite it as
dy = xcos(x^2)dx
Integrate both sides of the equation:
y = int xcos(x^2)dx
To take the integral on the right side of the equation, use the substitution method. Let z(x) = x^2 .
Then, dz = 2xdx and integral becomes
int xcos(x^2)dx = int(xdx) cos(x^2) = int 1/2 dz cosz = 1/2 int coszdz
This is a simple trigonometric integral: int cosz dz = sinz . Substituting the original variable, x, back into equation results in
y = 1/2 sinx^2 + C , where C is an arbitrary constant.
So, the general solution of the given differential equation is
y = 1/2sinx^2 + C .