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

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..

• Do you want to evaluate the integral in title or in the body? I don't see how they are the same...
– V.G
Feb 28, 2021 at 13:07
• Consider Leibniz rule for differentiating under integral sign Feb 28, 2021 at 13:07
• @LightYagami the body. Sorry I mistyped another problem!!! Feb 28, 2021 at 13:09
• @GerogeKlein do you know how to find the derivative of $$\int_{0}^{a(x)} \sin(\sqrt{t}) \mathrm{dt}$$ Feb 28, 2021 at 13:11

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.
• Congratulations for the $100k$ ! Feb 28, 2021 at 14:45