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DETAILS
Show that $G(x)=x\sin x+\cos x+C$ is the general antiderivative of $g(x)=x\cos x$.
$$\begin{array}{rl}\frac{d}{dx}(G(x))& =1\cdot ◻+x\cdot ◻-\sin x\\ & =◻+x-\sin x\\ & =x\cos x\end{array}$$

Step 1 ：Given the function $G(x)=x\sin x+\cos x+C$, we need to prove that it is the general antiderivative of $g(x)=x\cos x$.

Step 2 ：To do this, we take the derivative of $G(x)$ using the product rule and the chain rule.

Step 3 ：The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Step 4 ：The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 5 ：Applying these rules, we find that the derivative of $G(x)$ is indeed $g(x)=x\cos x$.

Step 6 ：This confirms that $G(x)$ is the antiderivative of $g(x)$.

Step 7 ：Final Answer: Therefore, we have shown that $G(x)=x\sin x+\cos x+C$ is the general antiderivative of $g(x)=x\cos x$. $\boxed{{\displaystyle G(x)=x\sin x+\cos x+C}}$