The equation of a curve passing through origin is given by $y=\int x^3\cos x^{4}dx$. f the equation of the curve is written in the form x = g(y), then

$\begin{array}{l}y=\int x^3\cos x^{4}dx\\ =\frac{1}{4}\int \cos tdt \left[Put{x}^4=t\Rightarrow x^{3}dx=\frac{1}{4}dt\right]\\ =\frac{1}{4}\sin t+C=\frac{1}{4}\sin x^4+C\end{array}$
$$\begin{gathered} \text{ Since the curve passes through }(0,0)\text{,therefore }C=0 \\ \begin{array}{l}\therefore y=\frac{1}{4}\sin x^4\\ \Rightarrow x^4={\sin }^{-1}\left(4y\right)\\ \Rightarrow x=\sqrt[4]{{\sin }^{-1}\left(4y\right)}\end{array} \end{gathered}$$