Question

In the following exercises, compute each definite integral. \(\int_{1 / 4}^{1 / 2} \frac{\tan \left(\cos ^{-1} t\right)}{\sqrt{1-t^{2}}} d t\)

Short answer

The definite integral evaluates to \(\ln(2)\).

Step-by-step solution

Understand the integral

The given integral is \(\int_{1/4}^{1/2} \frac{\tan(\cos^{-1} t)}{\sqrt{1-t^2}} \, dt\). This integral involves a trigonometric expression within the integrand.

Recognize the composition

Notice that \(\cos^{-1}(t) = \theta\) implies \(\cos(\theta) = t\). Hence, the expression \(\sqrt{1 - t^2}\) can be rewritten using \(\sin(\theta)\), as \(\sin^2(\theta) = 1 - \cos^2(\theta)\). Thus, \(\sqrt{1 - t^2} = \sin(\theta)\).

Simplify \(\tan(\cos^{-1} t)\)

Using that \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\), we can express \(\tan(\cos^{-1} t)\) as \(\frac{\sqrt{1-t^2}}{t}\).

Rewrite the integral

Substitute the expression for \(\tan(\cos^{-1} t)\) into the integral. It becomes \(\int_{1/4}^{1/2} \frac{\frac{\sqrt{1-t^2}}{t}}{\sqrt{1-t^2}} \, dt\), which simplifies to \(\int_{1/4}^{1/2} \frac{1}{t} \, dt\).

Evaluate the integral

Now, integrate \(\int \frac{1}{t} \, dt = \ln |t| + C\). Evaluate this definite integral from \(t = 1/4\) to \(t = 1/2\): \[ \left. \ln |t| \right|_{1/4}^{1/2} = \ln(1/2) - \ln(1/4). \]

Simplify the final expression

Use properties of logarithms to simplify: \[ \ln(1/2) - \ln(1/4) = \ln\left(\frac{1/2}{1/4}\right) = \ln(2). \] Thus, the value of the definite integral is \(\ln(2)\).

Key concepts

Trigonometric Identities

Trigonometric identities are essential tools in calculus. They help simplify complex expressions, especially when dealing with trigonometric functions like sine, cosine, and tangent. In this exercise, we encounter \(\tan(\cos^{-1} t)\), which involves combining inverse trigonometric functions.
Here's the crux: if \(\cos^{-1}(t) = \theta\), then the angle \(\theta\) has a \(\cos(\theta) = t\). This lets us convert expressions in terms of theta, the angle the original function refers to. Through the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\), we can express unknowns like \(\sin(\theta)\) as \(\sqrt{1 - t^2}\).
Keep in mind the tangent function defined as \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\). Substituting trigonometric identities transforms complex expressions into more manageable ones, enchanted by trigonometry's elegance.

Calculus

Calculus is the mathematical study of continuous change. It's part of the framework that allows us to understand changes in expressions, like the integral. The fundamental idea of calculus in this context is leveraging integration to transform complex algebraic and trigonometric expressions.
When addressing an integral, we must first understand the symbolic representation we're diving into — this isn't just fancy notation, but a compact, elegant description of sums over a range, here \([\frac{1}{4}, \frac{1}{2}]\).
Think of calculus as consisting of two main branches: differentiation and integration. Integration, focus here, accumulates quantities, akin to finding the total area under a curve given by a function. In our problem, breaking down the trigonometric complexity aids in calculating the integral effectively.

Integral Calculus

Integral calculus deals with the concept of summation; it's fundamentally about finding antiderivatives. In the context of definite integrals, it's about the calculation between specified limits to find the total value.
To compute the definite integral in this exercise, transforming \(\tan(\cos^{-1} t)/\sqrt{1-t^{2}}\)into \(1/t\) was key. The simplification involved using trigonometric identities to converge into a simpler integrable form.

• The recognition of \(\tan(\theta)\) allowed further simplification.
• The final integration required recognizing that the indefinite integral of \(1/t\) is \(\ln |t|\).

Upon finding this antiderivative, applying the limits \([1/4, 1/2]\) enables the evaluation, leading to the conclusion \(\ln(2)\), the definite integral's solution.