Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 18

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}
$