1. Evaluate the integral

Given

$$\lim_{x\to 0} \frac{\int_0^{x^2} \sec^2 t\,dt}{x\sin x}$$

We first compute the definite integral:

$$\int \sec^2 t\,dt = \tan t$$

So,

$$\int_0^{x^2} \sec^2 t\,dt = \tan(x^2)-\tan 0 = \tan(x^2)$$

Thus the limit becomes

$$\lim_{x\to 0} \frac{\tan(x^2)}{x\sin x}$$

2. Rewrite the expression

We use the standard limits near $0$:

$$\frac{\tan(x^2)}{x^2} \to 1, \qquad \frac{\sin x}{x} \to 1$$

Now,

$$\frac{\tan(x^2)}{x\sin x} = \frac{\tan(x^2)}{x^2}\cdot \frac{x}{\sin x}$$

3. Take the limit

As $x\to 0$,

$$\frac{\tan(x^2)}{x^2} \to 1$$

and

$$\frac{x}{\sin x} \to 1$$

Therefore,

$$\lim_{x\to 0} \frac{\tan(x^2)}{x\sin x} = 1\cdot 1 = 1$$

4. Check options

• A: $0$ ❌
• B: $3$ ❌
• C: $2$ ❌
• D: $1$ ✅

Hence the correct answer is D.