Question

10. Find $\frac{d^{2}}{d x^{2}}\int ^x_0\left(\int ^{\sin t}_1\sqrt{1+u^{4}}du\right)dt\text{.}$

Answer 1

Let g(t) denote the inner integral. By the fundamental theorem of calculus, the first derivative is

$\displaystyle \frac{d}{dx} \int_0^x g(t) \, dt = g(x)$

Then using the FTC again, differentiating g gives

$\displaystyle \frac{dg}{dx} = \frac{d}{dx} \int_1^{\sin(x)} \sqrt{1+u^4} \, du = \boxed{\cos(x) \sqrt{1+\sin^4(x)}}$