If $$f(x) = \int_0^x t(\sin x - \sin t)dt$$, then:

We start from the definition $$f(x)=\int_{0}^{x} t\bigl(\sin x-\sin t\bigr)\,dt.$$

To obtain the derivatives of $$f(x)$$ it is easier to differentiate under the integral sign rather than evaluate the integral outright. We recall Leibniz’s rule:

$$\frac{d}{dx}\int_{0}^{x}g(t,x)\,dt=g(x,x)+\int_{0}^{x}\frac{\partial g}{\partial x}(t,x)\,dt.$$

Here $$g(t,x)=t\bigl(\sin x-\sin t\bigr).$$

First we compute the two parts required by Leibniz’s rule.

Upper-limit term: $$g(x,x)=x\bigl(\sin x-\sin x\bigr)=0.$$

Partial-derivative term: $$\frac{\partial g}{\partial x}(t,x)=t\cos x$$ because $$\partial(\sin x)/\partial x=\cos x$$ while $$\sin t$$ is independent of $$x.$$ Hence

$$f'(x)=\int_{0}^{x}t\cos x\,dt.$$

The factor $$\cos x$$ is constant with respect to the dummy variable $$t$$, so

$$f'(x)=\cos x\int_{0}^{x}t\,dt =\cos x\;\frac{x^{2}}{2} =\frac{x^{2}}{2}\cos x.$$

Now we differentiate again to obtain $$f''(x).$$ Using the product rule $$\frac{d}{dx}\bigl(uv\bigr)=u'v+uv',$$ with $$u=\frac{x^{2}}{2}$$ and $$v=\cos x,$$ we get

$$f''(x)=\Bigl(\frac{x^{2}}{2}\Bigr)' \cos x+\frac{x^{2}}{2}\,(\cos x)' =x\cos x-\frac{x^{2}}{2}\sin x.$$

Next we differentiate once more to find $$f'''(x).$$ Again applying the product rule separately to the two terms of $$f''(x):$$

For $$x\cos x: \quad \frac{d}{dx}(x\cos x)=\cos x-x\sin x.$$ For $$-\dfrac{x^{2}}{2}\sin x: \quad \frac{d}{dx}\!\Bigl(-\frac{x^{2}}{2}\sin x\Bigr) =-\Bigl(\frac{x^{2}}{2}\Bigr)'\sin x-\frac{x^{2}}{2}\cos x =-x\sin x-\frac{x^{2}}{2}\cos x.$$

Adding these pieces gives

$$f'''(x)=\bigl(\cos x-x\sin x\bigr)+\bigl(-x\sin x-\frac{x^{2}}{2}\cos x\bigr) =\cos x-2x\sin x-\frac{x^{2}}{2}\cos x.$$

With $$f'(x)=\dfrac{x^{2}}{2}\cos x,$$ we now form the combination $$f'''(x)+f'(x)$$ that appears in Option D:

$$f'''(x)+f'(x) =\Bigl(\cos x-2x\sin x-\frac{x^{2}}{2}\cos x\Bigr)+\frac{x^{2}}{2}\cos x =\cos x-2x\sin x.$$

This result matches exactly the right-hand side given in Option D:

$$f'''(x)+f'(x)=\cos x-2x\sin x.$$

No other listed option yields an identity that agrees with our computations. Hence, the correct answer is Option 4.