${\dfrac{d}{dx}\Large\int} _{0}^{sinx} x^2\sqrt t\ \ dt$ as
${\dfrac{d}{dx}\ x^2 \Large\int} _{0}^{sinx} \sqrt t\ \ dt$
if so can i do the same thing for this too but I will endup with dt
${\dfrac{d}{dx}\Large\int} _{x}^{\sqrt x} \dfrac{e^x}{x}\ dt$
here all the variables are x except the dt can we pull out the $\dfrac{e^x}{x}$
${\dfrac{d}{dx}\dfrac{e^x}{x}\Large\int} _{x}^{\sqrt x} \ dt$
what can I do with ${\Large\int} _{x}^{\sqrt x} \ dt$?
thanks in advance
$dt$ is there to tell you with respect to what variable you're integrating. In your case, the variable of integration is $t$. Anything that's not $t$ is considered a constant (it's a number, in other words) that you can pull in and out of the integral sign. This is one of the most basic properties of both indefinite and definite integrals:
$$ \int af(x)\,dx=a\int f(x)\,dx $$
If it's $dt$, then it tells you that the variable of integration is $t$ and all other variables that are not $t$ are constants which means that they can be pulled out front. If it's $dx$, the variable of integration is $x$ and all other variables that are not $x$ are constants which, again, can be pulled out front. If it's $dy$, then your variable of integration is going to be $y$ and all other variables are considered constants and, sure enough, can be brought out of the integral sign. Long story short, $dX$ tells you that $X$ is the variable of integration which means you can't pull expressions containing $X$ out of the integral sign. Anything else, you can.
Take a look at this example:
$$ \int_{x}^{\sqrt{x}}x\,dt=x\int_{x}^{\sqrt{x}}\,dt=x\cdot t\bigg|_{x}^{\sqrt{x}}=x\left(\sqrt{x}-x\right)=x\sqrt{x}-x^2. $$
