91Ó°ÊÓ

A definite integral represents the area under a curve, giving us a number. It is written as \( \int_{a}^{b} f(t) \, dt \), where \(a\) and \(b\) are the limits of integration. This means we integrate the function \(f(t)\) from \(t = a\) to \(t = b\).

When working with definite integrals, the Fundamental Theorem of Calculus becomes crucial. It connects differentiation with integration, making it easy to find the derivative of an integral function.

In this exercise, we have \(F(x) = \int_{1}^{\cos x} \sqrt{1-t^2} \, dt\). The limits here are a constant, \(1\), and a variable expression, \(\cos x\). This setup requires careful application of the Fundamental Theorem with a twist because the upper limit is not simply \(x\).

To find the derivative of an integral with a variable limit, we apply the Fundamental Theorem of Calculus. When the upper limit \(b\) is a function of \(x\), such as \(\cos x\) in this problem, we must use the chain rule.

The Fundamental Theorem tells us:

• If \(F(x) = \int_{a}^{g(x)} f(t) \, dt\), then \(F'(x) = f(g(x)) \cdot g'(x)\).

In our original exercise:

• Function inside integrand: \(f(t) = \sqrt{1-t^2}\).
• Upper limit: \(\cos x\).
• Derivative of \(\cos x\): \(-\sin x\).

The result is \(F'(x) = -\sqrt{1-\cos^2{x}} = -\sqrt{\sin^2{x}}\). The negative sign arises from differentiating \(\cos x\) as part of applying the chain rule.

The Pythagorean identity is a fundamental trigonometric identity. It shows the relationship between the sine and cosine of angles:

\(\sin^2{x} + \cos^2{x} = 1\)

This identity helps to simplify expressions involving trigonometric functions. In our exercise, the expression \(\sqrt{1 - \cos^2{x}}\) appears.

By using the identity, we replace \(1 - \cos^2{x}\) with \(\sin^2{x}\). Thus, \(\sqrt{1 - \cos^2{x}}\) becomes \(\sqrt{\sin^2{x}}\), which simplifies to \(\sin{x}\).

This simplification is essential for finding the correct derivative of functions involving trigonometric square terms.