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The integral provided in the question is:

\[
\int_0^t \frac{\sin u}{u} \, du
\]

This is a standard form of the sine integral, also known as \(\text{Si}(x)\). The sine integral function \(\text{Si}(x)\) is given by:

\[
\text{Si}(x) = \int_0^x \frac{\sin u}{u} \, du
\]

Therefore, the integral in the question becomes:

\[
\text{Si}(t)
\]

Now, the provided options are expressed in terms of inverse trigonometric functions. This suggests the possibility of using an approximation or a result related to these forms. The correct result of this type of integral is commonly linked to the inverse tangent function. Specifically, for large \(t\), the sine integral can be approximated as:

\[
\int_0^t \frac{\sin u}{u} \, du \approx \frac{\pi}{2} - \frac{\cos(t)}{t}
\]

This form is related to the inverse tangent function, \(\tan^{-1}(t)\), for large values of \(t\). Looking at the options, option (a):

\[
\frac{\tan^{-1} s}{s}
\]

most closely resembles the standard approximation for this integral.

Thus, the correct answer is **(a)**.

Would you like more details on how this result was derived or explained further?

Here are 5 related questions you might find interesting:
1. What is the sine integral function \(\text{Si}(x)\)?
2. How is the approximation for \(\int_0^t \frac{\sin u}{u} \, du\) derived for large values of \(t\)?
3. What is the relationship between the sine integral and the inverse tangent function?
4. How do you evaluate definite integrals involving \(\sin(u)\) divided by \(u\)?
5. What other special functions are related to integrals of trigonometric functions?

**Tip:** When evaluating complex integrals, sometimes standard results or approximations involving known functions like \(\tan^{-1}(x)\) or \(\ln(x)\) are useful for simplifying the solution.

This problem involves evaluating the integral ∫₀ᵗ (sin u / u) du, which is a common type of integral encountered in advanced calculus. Learn how to solve it using the sine integral function Si(x) and explore related approximations.

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