Question

Find dy/dx:
y=x∫sin(t^3)dt

I didn't know how to put them in but the upper number on the integral symbol is x^2, and the lower number is 2.

Answer 1

\[x\int_2^{x^2}sin(t^3)dt\]

Answer 2

Thanks - do you know how to solve it? I was told to try it as a product rule but I'm not sure

Answer 3

I am not sure, let my try

Answer 4

it's not easy, but we have an outlet when you let u = t^3 , du = 3t^2 dt
since u = t^3 , t^2 = \(\Large u^{\frac{2}{3}}\) so, the integrand will become
\[x\int_8^{x^6} \frac{1}{3}u^{-2/3}sinu du\]
then take integral by part twice, you may get the answer