Find the derivative of the function $\displaystyle y = \int^0_{\frac{1}{x^2}} \sin^3 t dt$ using the 1st Fundamental Theorem of Calculus.

Using Properties of Integral

$\displaystyle \int^a_b f(x) dx = - \int^b_a f(x) dx$

So we have

$\displaystyle \int^0_{\frac{1}{x^2}} \sin^3 t dt = - \int^{\frac{1}{x^2}}_0 \sin ^ 3 t dt$

Let $\displaystyle u = \frac{1}{x^2}, \frac{du}{dx} = - \frac{2}{x^3}$. Then,

$ \begin{equation} \begin{aligned} \frac{d}{dx} \int^{\frac{1}{x^2}}_0 \sin^3 t dt =& \frac{d}{dx} \left(\int^u_0 - \sin^3 t dt \right) \\ \\ y' =& \frac{d}{du} \left(\int^u_0 - \sin ^3 t dt \right) \frac{du}{dx} \\ \\ y' =& (- \sin ^3 u) \frac{du}{dx} \\ \\ y' =& \left(- \sin ^3 \frac{1}{x^2}\right) \left( \frac{-2}{x^3} \right) \\ \\ y' =& \frac{2}{x^3} \sin ^2 \left( \frac{1}{x^2} \right) \end{aligned} \end{equation} $

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