91Ó°ÊÓ

The arcsecant, often written as \( \text{arcsec}(x) \), is one of the inverse trigonometric functions. It is the inverse of the secant function, meaning it helps us find an angle whose secant is \( x \). In mathematical terms, if \( y = \text{arcsec}(x) \), then \( \sec(y) = x \). This is important because the secant function is defined as \( \sec(\theta) = \frac{1}{\cos(\theta)} \), applicable where the cosine function is non-zero. When we encounter expressions like \( \text{arcsec}(x) \), it's useful to know these reverse relationships help in solving equations involving inverse trigonometric functions, as they provide a strategy to express angles in terms of trigonometric ratios. However, it's important to remember the arcsecant can only be applied to values \( x \geq 1 \) or \( x \leq -1 \), due to the nature of secant needing to be outside the range of \(-1, 1\) to avoid division by zero or undefined values.

Definite integrals are a fundamental concept in calculus. They represent the area under a curve on a graph within specified bounds, often denoted \( \int_{a}^{b} f(t) \, dt \). In this particular exercise, we have a definite integral that ranges from \( \sqrt{2} \) to \( x \), featuring the function \( \frac{1}{t \sqrt{t^2 - 1}} \). Understanding definite integrals requires grasping that they give a net area that the curve covers between the bounds. When applying inverse trigonometric functions, as shown here, we integrate to find that the antiderivative of \( \frac{1}{t \sqrt{t^2 - 1}} \) is \( \text{arcsec}(t) \). Thus, the definite integral \( \int_{\sqrt{2}}^{x} \frac{d t}{t \sqrt{t^2 - 1}} \) evaluates to \( \text{arcsec}(x) - \text{arcsec}(\sqrt{2}) \), allowing us to solve for \( x \) when given additional equation specifics like \( \frac{\pi}{2} \). By linking this integral with inverse trigonometric functions, complex integration becomes more manageable.

Solution verification is a crucial part of solving mathematical problems, ensuring your answer is correct. In this case, after finding \( x = -\sqrt{2} \), verifying involves checking whether the integration result satisfies the condition given in the exercise. For verification, replace \( x \) in the equation \( \text{arcsec}(x) - \text{arcsec}(\sqrt{2}) = \frac{\pi}{2} \). Calculate \( \text{arcsec}(-\sqrt{2}) \) and \( \text{arcsec}(\sqrt{2}) \). We've found that \( \text{arcsec}(\sqrt{2}) = \frac{\pi}{4} \), and earlier calculated \( \text{arcsec}(-\sqrt{2}) = \frac{3\pi}{4} \). Putting these into the expression gives \( \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2} \), exactly as required. Verification confirms the computations were done correctly and the integral condition holds true, reinforcing the reliability of your solution.