Answer
The co-efficient of $x^3$ is $\dfrac{1}{3}$and the co-efficient of $x^4$ is $0$.
Work Step by Step
Write the Taylor series for the function as: $\sin x=x-\dfrac{(x)^3}{3!}+\dfrac{(x)^5}{5!}+.......=\Sigma_{k=0}^{\infty} (-1)^k \dfrac{x^{2k+1}}{(2k+1)!}$
Replace $x$ by $t^2$ in the above series to obtain:
$\int_{0}^{x} \sin (t^2) \ dt=\int_0^x[(t^2)-\dfrac{(t^2)^3}{3!}+\dfrac{(t^2)^5}{3!}-........]\ dt \\=[\dfrac{t^3}{3}-\dfrac{t^7}{(7)(3!)}+\dfrac{t^{11}}{(11)(5!)}-......]_0^x\\=\dfrac{x^3}{3}-\dfrac{x^7}{(7)(3!)}+\dfrac{x^{11}}{(11)(5!)}-......$
Therefore, we find that the co-efficient of $x^3$ is $\dfrac{1}{3}$and the co-efficient of $x^4$ is $0$.
