How to find derivative of $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ I am asked to solve this question:
'Which one is the higher order infinitesimals when $\mathbf{x \rightarrow 0 \, \int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}$' or $\mathbf{x^2}$.
I know that:I have to solve $\lim_{x \rightarrow 0}\frac{\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt}{x}$ by using L 'Hospital's rule and compare the result with $x^2$
But I cant find the derivative.
I have recited that the derivative of $\int_{φ(x)}^{ψ(x)}f(t)dt$ (if x is not in f(t)) is $f[φ(x)]φ'(x)-f[(ψ)]ψ'(x)$, without actually knowing how it comes.
But for now, this equation seems impossible since I cannot get $(1+0)^{\frac{1}{0}}$, I don't think $\lim (1+0)^{\frac{1}{0}}$ should be right, since the hint said the $\int_0^{\sin x} (1+t)^{\frac{1}{t}} dt$ ~ $ex$
So what should I do? Is the equation I have recited succeed all the time?
 A: First note that, as $t\to 0$,
$$
(1 + t)^{\frac{1}{t}}  = \exp \left( {\frac{1}{t}\log (1 + t)} \right) = \exp \left( {1 + \mathcal{O}(t)} \right) = e\exp \left( {\mathcal{O}(t)} \right) = e + \mathcal{O}(t).
$$
Hence
$$
\int_0^{\sin x} {(1 + t)^{\frac{1}{t}} dt}  = e\sin x + \mathcal{O}(\sin ^2 x) = ex + \mathcal{O}(x^2 )
$$
as $x\to 0$.
A: I expand on my comment for your benefit.
Let's first analyze the integral $\int_0^{\sin x} (1+t)^{1/t}\,dt$. You should note that the integrand is not defined at $t=0$. However, by definition the symbol $\int_a^b f(x) \, dx$ requires $f$ to be defined in whole interval $[a, b] $.
In such a situation when the integrand is not defined at a finite number of points in the interval of integration, the  convention is to define the function at those exceptional points in any manner whatsoever. This does not create any problem because the value of an integral or its existence is not affected by behavior of function at a finite number of points.
Next we define the integrand in question at $t=0$ in such a manner that the function becomes continuous there so that we can take advantage of the Fundamental Theorem of Calculus.
Let $f(0)=e$ and $f(t)=(1+t)^{1/t}$ for $t\neq 0$. Then $f$ is defined in some neighborhood of $0$ and moreover is continuous in that neighborhood.
Now $$\lim _{x\to 0}\frac{1}{x}\int_0^{\sin x} (1+t)^{1/t}\,dt=\lim_{x\to 0}\frac{\sin x}{x}\cdot\frac{1}{\sin x} \int_{0}^{\sin x} f(t) \, dt$$ The first fraction tends to $1$ and next we replace $\sin x$ with $u$ and note that $u$ tends to $0$ with $x$. Hence the desired limit is $$\lim_{u\to 0}\frac{1}{u}\int_0^u f(t) \, dt$$ By FTC this equals $f(0)=e$ as $f$ is continuous at $0$.
