问题及解答

$F(x)=\sin x^2\cdot\int_0^1f(t\sin x^2)dt$, 求 $\frac{dF}{dx}$.

令 $u=t\sin x^2$, 则 $t=\frac{u}{\sin x^2}$, $dt=\frac{1}{\sin x^2}du$. 于是

\[
F(x)=\sin x^2\cdot\int_0^{\sin x^2}f(u)\cdot\frac{1}{\sin x^2}du=\int_0^{\sin x^2}f(u)du,
\]

从而

\[
\frac{dF}{dx}=f(\sin x^2)\cdot\cos x^2\cdot 2x=2x\cos x^2\cdot f(\sin x^2).
\]