# Using L'Hospital rule, how do we tell that $\int_0^x \sin^{-1} (t^2) dt =0$?

Find the limit, $$\lim_{x \to 0+} \frac{\int_0^x \sin^{-1} t^2 dt}{\sin(x^2)}$$

This came out in my exam and unknowingly, I did the L'Hospital rule and got it correct, but I do not understand how $$\int_0^x \sin^{-1} t^2 dt = 0$$ ?

The denominator is $$0$$ as $$\sin (0) =0$$ and I used LH rule because it is in the form of $$\frac{0}{0}$$

Using L-H rule on the numerator, it involves the first fundamental theroem of calculus

$$\frac{d}{dx} \int_0^x \sin^{-1} t^2 dt = \sin^{-1} x^2$$ and I would know how to carry on. So, how do I determine that $$\int_0^x \sin^{-1} t^2 dt = 0$$ ?

• Why do you want to "determine that $\int_0^x \sin^{-1} t^2 dt = 0$" (whatever that means)? You solved the exercise without this, didn't you? Commented Nov 20, 2022 at 8:43
• The numerator tends to $0$ because $\sin^{-1}$ is bounded. Commented Nov 20, 2022 at 8:44
• Or simply by continuity of $x\mapsto\int_0^x\dots$ (which is even differentiable, as you showed). Commented Nov 20, 2022 at 8:49

When you plug in $$0$$ for the limit you’re plugging in $$0$$ for $$x$$ not $$t$$.

$$\lim_{x \to 0+} \frac{\int_0^x \sin^{-1} t^2 dt}{\sin(x^2)}$$ basically means $$\frac{\int_0^0\sin^{-1} t^2 dt}{\sin(0^2)}=0/0$$

Which is obvious since an integral with the same bounds will always equal $$0$$. Because it’s $$0/0$$ then you’d apply le hospital rule and so on the solve the limit.

This means that $$\int_0^x \sin^{-1} t^2 dt\neq 0$$, but $$\int_0^0\sin^{-1} t^2 dt=0$$

Hopefully this clears up some misunderstanding.

As a side note the actual integral can be expressed in terms of hypergeometric functions as shown here https://www.wolframalpha.com/input?key=&i=integrate+arcsin%28x%5E2%29+from+0+to+x

Let $$N(x):=\int_0^x \sin^{-1} (t^2)\,\mathrm dt.$$

As you showed, $$N$$ is differentiable. Hence it is continuous, so that $$\lim_{x\to0}N(x)=N(0)=0,$$ which is the only thing you missed to apply L'Hôpital's rule.