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I encountered the following question:

Find, $$ \frac{d^2}{dx^2}\int_{0}^{x}\left(\int_{1}^{\sin(t)}\sqrt{1+u^4}\,du\right)dt $$

I attempted to evaluate the above using FTC twice, however, I don't think that purely applying FTC twice is a legal method since the derivative is twice with respect to $x$.

Any help is much appreciated!

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The first derivative indeed works with FTC. Do that, now we've got: $$\frac{d}{dx}\int_1^{\sin(x)}\sqrt{1 + u^4} du$$ Let $f(u)$ be an anti-derivative of $\sqrt{1+u^4}$, such that $f'(u) = \sqrt{1+u^4}$. Then this integral can be written as: $$\frac{d}{dx}\int_1^{\sin(x)}f'(u) du$$ $$=\frac{d}{dx}[f(\sin(x)) - f(1)]$$ $$=f'(\sin(x))\cos(x)$$ $$=\cos(x)\sqrt{1+\sin^4(x)}$$ The above method works for any "FTC-like" problems where the upper and lower bounds are part of the derivative.

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