Jtdylktuy

$$sin xfrac {dy}{dx}+(cos x)y=sin(x^2)$$
$$frac {d}{dx} y sin x=sin(x^2)$$
$$ysin x=int sin(x^2)dx = -frac{1}{2x}cos(x^2)+C$$
$$y=-frac{cos(x^2)}{2xsin x}+frac {C}{sin x}$$
where C is constant

Is my answer correct?

The answer given in your textbook is correct - you have tried to oversimplify it! You got to $$frac {d}{dx} y sin x=sin(x^2)$$Then you integrate both sides $$ysin x=intsin(x^2) mathrm dx+C$$Then you divide by $sin x$ to get $$y=frac{intsin(x^2)mathrm dx+C}{sin x}$$

as required.

There is no need to try and integrate the $sin(x^2)$ term, the way you did it is not correct. To check why this is the case, try to differentiate your result. If you had done it correctly, it would give you $sin(x^2)$. However, what it really gives you is $$sin(x^2)+frac{cos(x^2)}{2x^2}$$so there has been a mistake.

The obeservation you have done is pretty good! You have correctly identified the LHS as an application of the product rule since

$$frac{mathrm d}{mathrm d x}left(ysin(x)right)=frac{mathrm d y}{mathrm d x}sin(x)+ycos(x)$$

However, without trying to discourage you but the integration of $sin(x^2)$ is sadly speaking not that simple. You can check that you conjectured anti-derivative is wrong by simple taking the derivative. Thus, the whole solution is given by the following

$$begin{align*}
frac{mathrm d y}{mathrm d x}sin(x)+ycos(x)&=sin(x^2)\
frac{mathrm d}{mathrm d x}left(ysin(x)right)&=sin(x^2)\
ysin(x)&=intsin(x^2)mathrm d x+C
end{align*}$$

$$therefore~y(x)~=~frac1{sin(x)}left(intsin(x^2)mathrm d x+Cright)$$

Adding some details concerning the still remaining integral: According to WolframAlpha the integral can be written in terms of the special function Fresnel Integral. For a straightforward "solution" you can expand the sine as a series and integrate termwise. On the other hand for definite integrals there are some value known, the I would say most important is

$$int_0^infty sin(x^2)mathrm d x~=~frac{sqrt{pi}}{2sqrt 2}$$