1. Given function

We have

$$f(x)=\int_0^x t(\sin x-\sin t)\,dt.$$

We want to test which differential relation is satisfied by $f(x)$.

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2. Rewrite the integral

Since $\sin x$ is constant with respect to $t$,

$$f(x)=\sin x\int_0^x t\,dt-\int_0^x t\sin t\,dt.$$

Now,

$$\int_0^x t\,dt=\frac{x^2}{2}.$$

So,

$$f(x)=\frac{x^2}{2}\sin x-\int_0^x t\sin t\,dt.$$

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3. Differentiate using Leibniz rule

A cleaner way is to differentiate the original form directly.

Let

$$F(x,t)=t(\sin x-\sin t).$$

Then

$$f(x)=\int_0^x F(x,t)\,dt.$$

Using Leibniz rule,

$$f'(x)=\int_0^x \frac{\partial F}{\partial x}(x,t)\,dt+F(x,x).$$

Now,

$$\frac{\partial F}{\partial x}=t\cos x,$$

and

$$F(x,x)=x(\sin x-\sin x)=0.$$

Hence,

$$f'(x)=\int_0^x t\cos x\,dt=\cos x\int_0^x t\,dt=\frac{x^2}{2}\cos x.$$

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4. Second derivative

Differentiate:

$$f''(x)=\frac{d}{dx}\left(\frac{x^2}{2}\cos x\right).$$

Using product rule,

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

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5. Third derivative

Differentiate again:

$$f'''(x)=\frac{d}{dx}\left(x\cos x-\frac{x^2}{2}\sin x\right).$$

Now,

$$\frac{d}{dx}(x\cos x)=\cos x-x\sin x,$$ $$\frac{d}{dx}\left(\frac{x^2}{2}\sin x\right)=x\sin x+\frac{x^2}{2}\cos x.$$

Therefore,

$$f'''(x)=\cos x-x\sin x-\left(x\sin x+\frac{x^2}{2}\cos x\right)$$

so

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

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6. Check each option

We already have

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

Option A

$$f'''(x)+f''(x)$$

contains terms like $x\cos x$, $x^2\sin x$, etc., so it cannot simplify to just $\sin x$.

Hence, A is false.

Option B

$$f'''(x)+f''(x)-f'(x)$$

also contains extra $x$ and $x^2$ terms, so it is not equal to just $\cos x$.

Hence, B is false.

Option C

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

The $\frac{x^2}{2}\cos x$ terms cancel, giving

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

This matches exactly.

Hence, C is true.

Option D

$$f'''(x)-f''(x)$$

will still contain extra terms and does not reduce to $\cos x-2x\sin x$.

Hence, D is false.

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7. Final answer

The correct option is

$$\boxed{\text{C}}$$

with relation

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

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8. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.