Problem 4-25 in Spivak's book

[kakarotyjn]

Let c be a singular k-cube and $$p:[0,1]^k \to [0,1]^k$$ a 1-1 function such that $$p([0,1]^k ) = [0,1]^k$$ and $$\operatorname{det} p'(x) \ge 0$$ for $$x \in [0,1]^k$$.If $$\omega$$ is a k-form,show that $$\int\limits_c \omega = \int\limits_{c \circ p} \omega$$

Note that
$$\int\limits_c \omega = \int\limits_{[0,1]^k } {c*\omega } = \int\limits_{[0,1]^k } {(f \circ c)(\det c')dx^1 \wedge ... \wedge dx^k }$$
$$\int\limits_{c \circ p} \omega = \int\limits_{[0,1]^k } {(c \circ p)*\omega } = \int\limits_{[0,1]^k } {(f \circ c \circ p)(\det (c \circ p)')dx^1 \wedge ... \wedge dx^k } = \int\limits_{[0,1]^k } {(f \circ c \circ p)((\det c') \cdot (\det p'))dx^1 \wedge ... \wedge dx^k }$$

did I deduce it right?If it's right,how to prove

 

[fresh_42]

Looks good, but you have to apply the transformation theorem on $p$ as the final step.