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Compute ∫1 -1 (i don't know how to write this properly in open study) 2x cos(x2)dx
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\[\int 1-2x\cos(x^2)\,dx\] is this your question?
if so, use the substitution\[u=x^2\] then, \[du=2xdx\] now back to the integral \[\int 1-2x\cos(x^2)\,dx=\int\,dx-\int \cos(u)\,du=x-\sin(u)+c=x-\sin(x^2)+c \]
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