>>865
cの解答
∫[x,x^2]cos(t^2)dt
=∫[0,x^2]cos(t^2)dt-∫[0,x]cos(t^2)dt

ここで
d/dx(∫[0,x]cos(t^2)dt)
=cos(x^2)

またx^2=uとおくと
d/dx(∫[0,x^2]cos(t^2)dt)
=(du/dx)･(d/du(∫[0,u]cos(t^2)dt))
=2x･cos(x^4)

と言うことで
d/dx(∫[x,x^2]cos(t^2)dt)
=2x･cos(x^4)-cos(x^2)