Question

Perform the indicated integrations. $$ \int \frac{d t}{2 t \sqrt{4 t^{2}-1}} $$

Short answer

\( \int \frac{d t}{2 t \sqrt{4 t^{2}-1}} = \frac{1}{2} \sec^{-1}(2t) + C \)

Step-by-step solution

Identify the substitution

Notice that the expression under the square root in the denominator resembles a typical form that can be simplified using a trigonometric substitution. Here, we recognize that the term is in the form of \( a^2t^2 - b^2 \), which suggests a substitution of \( t = \frac{1}{2} \sec(\theta) \).

Perform the substitution

Substitute \( t = \frac{1}{2} \sec(\theta) \). This implies \( dt = \frac{1}{2} \sec(\theta)\tan(\theta) \, d\theta \). Substitute these into the integral:\[\int \frac{\frac{1}{2} \sec(\theta)\tan(\theta) \, d\theta}{2 \left(\frac{1}{2} \sec(\theta)\right) \sqrt{4 \left(\frac{1}{2} \sec(\theta)\right)^{2} - 1}}\]

Simplify the expression

Substitute and simplify inside the integral:\[\int \frac{\frac{1}{2} \sec(\theta)\tan(\theta) \, d\theta}{\sec(\theta) \sqrt{\sec^2(\theta) - 1}}\]Since \( \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \), further simplification gives:\[\int \frac{\frac{1}{2} \sec(\theta)\tan(\theta) \, d\theta}{\sec(\theta) \tan(\theta)} = \int \frac{1}{2} \, d\theta\]

Integrate with respect to θ

The integral \( \int \frac{1}{2} \, d\theta \) is straightforward:\[ \frac{1}{2} \int d\theta = \frac{1}{2} \theta + C \]where \( C \) is the constant of integration.

Substitute back to the original variable

To find \( \theta \) in terms of \( t \), recall the substitution \( t = \frac{1}{2} \sec(\theta) \), which means \( \sec(\theta) = 2t \) and thus \( \theta = \sec^{-1}(2t) \). Substitute this back:\[ \frac{1}{2} \theta = \frac{1}{2} \sec^{-1}(2t) + C \]. Consequently, the integrated function in terms of \( t \) is \[ \frac{1}{2} \sec^{-1}(2t) + C \].

Key concepts

Integration Techniques

When tackling integrals, especially ones that look complex, understanding various integration techniques becomes crucial. One common technique is **trigonometric substitution**. This involves replacing certain expressions within the integral with trigonometric identities. For integrals involving functions like \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{x^2 + a^2} \), we can often simplify calculations by substituting with sine, cosine, or secant functions.
In our exercise, we use the substitution \( t = \frac{1}{2} \sec(\theta) \) to simplify the square root expression in the denominator. With this substitution, we convert the integral into one that's more manageable by transforming the variables from 't' to 'θ'. By doing so, the integration process often becomes more direct and allows for easier application of standard integration formulas. Consistently practicing these techniques sharpens intuition for recognizing suitable substitutions and simplifications.

To excel at using these techniques, be consistent with practicing different problems. Understand the distinct forms that can be resolved with trigonometric identities and always double-check the variable changes throughout the process.

Definite Integrals

Definite integrals provide not just antiderivatives but numerical values representing areas under curves. They are expressed in the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the bounds of integration. This is unlike indefinite integrals, which include a constant of integration and represent a family of functions.

While our original exercise is an indefinite integral, it's worthwhile to consider how definite integration would apply. If the bounds were specified, after performing the trigonometric substitution and simplification, you'd evaluate the resulting antiderivative at these bounds. This often involves substituting back to the original variable post-integration, then calculating the difference as per the Fundamental Theorem of Calculus. Practicing definite integrals enhances one's skill in applying boundaries to ensure correct problem-solving execution in real-world scenarios where you'd need exact computations.

The key takeaway here is understanding that, in definite integrals, simplifying before applying bounds can often simplify calculations and lead to more precise solutions.

Inverse Trigonometric Functions

Inverse trigonometric functions come into play often when reversing trigonometric transformations. In the context of integration, they are essential when changing variables back to the original form. In the example from the original exercise, after substituting \( t = \frac{1}{2} \sec(\theta) \), the resulting solution \( \frac{1}{2} \theta \) needs to be expressed back in terms of 't'.

To achieve this, inverse trigonometric functions contextualize the angle in terms of the original variable. So, \( \theta = \sec^{-1}(2t) \) allows us to rewrite the integrated function in terms of 't' as \( \frac{1}{2} \sec^{-1}(2t) + C \). Understanding these functions and their properties—domains, ranges, and the typical transformations they involve—is crucial for not only simplifying integrals but also in solving equations and representing real-world scenarios accurately.

Regular practice with inverse trigonometric functions will make it easier to switch between inverse and direct trig functions seamlessly, an important skill for problem-solving in calculus and beyond.