Answer
$$g'(x) = 2(1-2x) \sin (1-2x)+2(1+2x) \sin (1+2x)$$
Work Step by Step
Given
\begin{aligned} g(x)=\int_{1-2 x}^{1+2 x} t \sin t d t\end{aligned}
Since
\begin{aligned}
g(x)&=\int_{1-2 x}^{1+2 x} t \sin t d t\\
&=\int_{1-2 x}^{0} t \sin t d t+\int_{0}^{1+2 x} t \sin t d t
\end{aligned}
Then by using Fundamental Theorem of calculus
\begin{aligned}
g'(x) &=\frac{d}{dx}\int_{1-2 x}^{0} t \sin t d t+\frac{d}{dx}\int_{0}^{1+2 x} t \sin t d t\\
&=-\frac{d}{dx}\int_{0}^{1-2 x} t \sin t d t+\frac{d}{dx}\int_{0}^{1+2 x} t \sin t d t\\
&= -\left( (1-2x) \sin (1-2x)\right) \frac{d}{dx}\left( (1-2x) \right) +
\left( (1+2x) \sin (1+2x)\right) \frac{d}{dx}\left( (1+2x) \right)\\
&= 2(1-2x) \sin (1-2x)+2(1+2x) \sin (1+2x)
\end{aligned}
