1. Given function

$$g(x)=\int_0^x \cos 4t\,dt$$

Let us first evaluate this integral.

$$g(x)=\left[\frac{\sin 4t}{4}\right]_0^x=\frac{\sin 4x}{4}$$

So,

$$g(x)=\frac{\sin 4x}{4}$$

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2. Find $g(x+\pi)$

Substitute $x+\pi$ in place of $x$:

$$g(x+\pi)=\frac{\sin 4(x+\pi)}{4}$$ $$=\frac{\sin(4x+4\pi)}{4}$$

Using periodicity of sine,

$$\sin(4x+4\pi)=\sin 4x$$

Hence,

$$g(x+\pi)=\frac{\sin 4x}{4}=g(x)$$

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3. Now compute $g(\pi)$

$$g(\pi)=\int_0^{\pi}\cos 4t\,dt=\left[\frac{\sin 4t}{4}\right]_0^{\pi}$$ $$=\frac{\sin 4\pi-\sin 0}{4}=0$$

So,

$$g(\pi)=0$$

Therefore,

$$g(x)+g(\pi)=g(x)+0=g(x)$$

and

$$g(x)-g(\pi)=g(x)-0=g(x)$$

Since we already found that

$$g(x+\pi)=g(x)$$

it follows that

$$g(x+\pi)=g(x)+g(\pi)$$

and also

$$g(x+\pi)=g(x)-g(\pi)$$

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4. Check all options

• A: $\dfrac{g(x)}{8\pi}$

  This is not equal to $g(x)$ in general. So incorrect.

• B: $g(x)+g(\pi)$

  Since $g(\pi)=0$, this equals $g(x)$. So correct.

• C: $g(x)-g(\pi)$

  Since $g(\pi)=0$, this also equals $g(x)$. So correct.

• D: $g(x)\cdot g(\pi)$

  Since $g(\pi)=0$, this becomes $0$, not equal to $g(x)$ in general. So incorrect.

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5. Final answer

The correct options are:

$$\boxed{\text{B, C}}$$