Calculus: Early Transcendentals, Chapter 5, 5.5, Section 5.5, Problem 7

You need to evaluate the indefinite integral by performing the substitution x^2 = t , such that:
x^2 = t => 2xdx = dt => xdx = (dt)/2
int x*sin(x^2) dx= (1/2)int sin t dt
(1/2)int sin t dt= (1/2)(-cos t) + c

Replacing back x^2 for t yields:
int x*sin(x^2) dx= -(1/2)(cos x^2) + c
Hence, evaluating the indefinite integral yields int x*sin(x^2) dx= -(1/2)(cos x^2) + c