Prove $\frac{d}{dx} \int_x^{x^2}\ \frac{\sin t}{t} dt = \frac{2\sin x^2 - \sin x}{x}$

Question

Check whether the following is true:

$$\frac{d}{dx} \int_x^{x^2}\ \frac{\sin t}{t} dt = \frac{2\sin x^2 - \sin x}{x}$$ .

If not true then prove it wrong.

I know how to evaluate $$\int\frac{\sin t}{t} dt$$

First we can write the Tyalor expansion of $\sin t$ and then further integrating. But it will be a long tedious. I want to know some easy solution to this particular problem.

At first I was trying to evaluate $$\int \frac{sin t}{t} dt$$ using by parts it was again and again redirected to the original integral.

After trying again and again I'm unable to come to a proper final result. Kindly help.

Answer 1

Hint

The fundamental theorem of calculus write $$\frac d {dx}\int_{a(x)}^{b(x)} f(t) \, dt=f\big(b(x)\big)\, b'(x)-f\big(a(x)\big)\, a'(x)$$ One term disappears when one of the bounds is a constant.

Answer 2

HINT: use the fact that $$\frac{d}{dx}\int_c^{g(x)}f(t)dt=f(g(x))\cdot g'(x)$$

where $c$ is some constant.