$f(x) = \int\limits_{a(x)}^{b(x)} \sin{\sqrt t}\,\mathrm{d}t,$ compute $f^{'}(0)$.

Question

Let $a(x)=\frac{\pi^2}{4} + \cos(3x+\frac{\pi}{2})$ and $b(x)=\frac{25 \pi^2}{4} + 2x^2$. $$f(x) = \int\limits_{a(x)}^{b(x)} \sin{\sqrt t}\,\mathrm{d}t,$$ compute $f^{'}(0)$.

My thoughts:

To evaluate $$\int \sin \sqrt x \, dx $$, let $u = \sqrt x, x = u^2, dx = 2u \, du$

$$\int \sin \sqrt x \, dx = \int \sin u 2u \, du = 2 \int u d (-\cos u) \, du = 2 \left( -u\cos u + \int \cos u \, du \right) = 2 \left( -u\cos u + \sin u + C \right) $$

But it doesn't make the final computation much easier..

Answer 1

More generally, if $f$ has an antiderivative $F$,$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}[F(b(x))-F(a(x))]=f(b(x))b^\prime(x)-f(a(x))a^\prime(x),$$so we don't need to know $F$ itself, i.e. you don't need to integrate $f$. (Rather nicely, you don't need to differentiate $f$ either.) Now take$$a=\frac{\pi^2}{4}-\sin(3x),\,b=\frac{25\pi^2}{4}+2x^2,\,f=\sin\sqrt{t}.$$I'll leave that to you.