Answer
$-\dfrac{1}{2} \cos (x^2-2x+1)++C$
Work Step by Step
Our aim is to solve the integral $ \int (x-1) \sin (x^2-2x+1)\ dx$
Let us consider that $a =x^2-2x+1$ and $\dfrac{da}{dx}=(2x-2) =2(x-1)$
Now, $\int (x-1) \sin (x^2-2x+1)\ dx = \int (x-1) \sin a \times [\dfrac{1}{2(x-1)}] \ da $
or, $=\dfrac{1}{2} \int \sin a da $
or, $=-\dfrac{1}{2} \cos a+C$
or, $=-\dfrac{1}{2} \cos (x^2-2x+1)++C$
