# How to you find the general solution of dy/dx=xcosx^2?

How to you find the general solution of $\frac{dy}{dx}=x{\mathrm{cos}x}^{2}$?
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Maggie Tanner
First we notice that
$d\frac{{\mathrm{sin}x}^{2}}{dx}=2x{\mathrm{cos}x}^{2}$
or
$x{\mathrm{cos}x}^{2}=\frac{1}{2}\cdot \left(d\frac{{\mathrm{sin}x}^{2}}{dx}\right)$
Hence the problem becomes
$\frac{dy}{dx}=\frac{1}{2}\cdot \frac{d{\mathrm{sin}x}^{2}}{dx}$
Integrate both sides with respect to x and we have
$\int \frac{dy}{dx}\cdot dx=\frac{1}{2}\cdot \int \left(d\frac{{\mathrm{sin}x}^{2}}{dx}\right)dx$
$y=\frac{1}{2}\cdot {\mathrm{sin}x}^{2}+c$
The general solution is
$y\left(x\right)=\frac{1}{2}\cdot {\mathrm{sin}x}^{2}+c$

iescabroussexg
Another way
$\frac{dy}{dx}=x{\mathrm{cos}x}^{2}$
$dy=x{\mathrm{cos}x}^{2}dx$
$\int dy=\int x{\mathrm{cos}x}^{2}dx$
THIS SOLUTION IS ONLY CORRECT IF THE PROBLEM IS WRITTEN CORRECTLY. The solution would be different if the problem is $\frac{dy}{dx}=x{\mathrm{cos}}^{2}x$.
Let $u={x}^{2}$. Then du=2xdx and $dx=\frac{du}{2x}$
$\int dy=\int x\mathrm{cos}u\frac{du}{2x}$
$\int dy=\int \frac{1}{2}\mathrm{cos}udu$
$y=\frac{1}{2}\mathrm{sin}u+C$
$y=\frac{1}{2}\mathrm{sin}\left({x}^{2}\right)+C$