Problem 31 - Thomas Calculus

Question

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int_{1 / 12}^{1 / 4} \frac{2 d t}{\sqrt{t}+4 t \sqrt{t}} $$

Step-by-Step Solution

Answer

Apply substitution and evaluate integration to find its value.

Step 1: Simplify the Integrand

We begin with the integral: \[ \int_{1/12}^{1/4} \frac{2}{\sqrt{t} + 4t\sqrt{t}} \, dt \] Notice that \( \sqrt{t} = t^{1/2} \), thus the integrand can be rewritten as:\[ \frac{2}{t^{1/2} + 4t^{3/2}} \]Simplifying further:\[ \frac{2}{t^{1/2}(1 + 4t)} \] Now our integral is:\[ \int_{1/12}^{1/4} \frac{2}{t^{1/2}(1 + 4t)} \, dt \]

Step 2: Substitute for Simplified Form

Let's use the substitution:\[ u = \sqrt{t} \rightarrow t=u^2 \text{ and } dt = 2u \, du \]The limits of integration change as well. When \( t = \frac{1}{12} \), \( u = \sqrt{\frac{1}{12}} \), and when \( t = \frac{1}{4} \), \( u = \frac{1}{2} \).The integral becomes:\[ \int_{\sqrt{1/12}}^{1/2} \frac{2}{u(1+4u^2)} \, (2u \, du) \]\[= \int_{\sqrt{1/12}}^{1/2} \frac{4}{1+4u^2} \, du \]

Step 3: Perform a Trigonometric Substitution

For \( \int \frac{1}{1+4u^2} \, du \), we use the substitution:\[ 2u = \tan \theta \rightarrow du = \frac{1}{2} \sec^2 \theta \, d\theta \]The integral becomes:\[ \int \frac{4}{1+\tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta \, d\theta \]\[ = \int 2 \cos^2 \theta \, d\theta \]\[ = 2 \int (\frac{1+\cos 2\theta}{2}) \ d\theta \]\[ = \int (1 + \cos 2\theta) \, d\theta \]

Step 4: Solve the Integral

Calculate the integral:\[ \int (1 + \cos 2\theta) \, d\theta = \theta + \frac{1}{2} \sin 2\theta + C \]

Step 5: Reverse Substitution

Substitute back to the original variable \( u \) and \( t \):Since \( 2u = \tan \theta \), \( \theta = \tan^{-1}(2u) \).Now calculate the limits or evaluate the definite integral back from \( \theta \).

Step 6: Evaluate Definite Integral

Substitute the \( u \) limits back into \( \theta = \tan^{-1}(2u) \) expressions and solve:Compute \[ \theta \bigg|_{\sqrt{1/12}}^{1/2} \] and convert back if needed to find the definite integral result.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique often used in calculus to simplify certain integrals, especially those involving square roots. By substituting a trigonometric function for a variable, complex integrals can become simpler to evaluate.
In our exercise, the integrand has the form of an expression that matches a known trigonometric identity: the sum of squares. By using the substitution \( 2u = \tan \theta \), the denominator expression \( 1 + 4u^2 \) becomes \( 1 + \tan^2\theta \), which simplfies to \( \sec^2\theta \).

• Trigonometric identities, like \(1 + \tan^2\theta = \sec^2\theta\), help in reducing expressions to simpler forms.
• Once the substitution is made, the trigonometric identity can transform the problem into one involving much simpler functions, like \( \cos^2 \theta \).

Understanding these conversions is crucial, as they greatly aid in evaluating otherwise tricky integrals.

Integral Evaluation

The process of integral evaluation involves finding the antiderivative, or the original function, for the given integrand. This is often necessary for understanding the entire function represented by the integral.
In our exercise, the integral was simplified using trigonometric identities to \( \int (1 + \cos 2\theta) \, d\theta \). This step is a direct consequence of recognizing and simplifying using trigonometric identities.

• The integral of \(1\) with respect to \(\theta\) is straightforward: \(\theta\).
• For \(\cos 2\theta\), the antiderivative is \(\frac{1}{2}\sin 2\theta\).

After solving these, by adding a constant of integration \(C\), the evaluated integral becomes \(\theta + \frac{1}{2} \sin 2\theta + C\). This expression is crucial for back-substituting to reconstruct our original variable terms.

U-Substitution

U-substitution is a technique used to simplify the integral before proceeding to more complex methods. It involves substituting part of the integrand with a new variable, \(u\), making the integration process more straightforward.
In this case, by choosing \(u = \sqrt{t} \), we simplify the expression \(4t\sqrt{t}\) to \(4u^3\). This simplification allowed us to adjust the integral limits and rewrite the integral in terms of \(u\), creating the integral \( \int \frac{4}{1+4u^2} \, du \).

• When replacing \(t\) with \(u\), remember to also convert \(dt\) accordingly: if \(t = u^2\), then \(dt = 2u \, du\).
• Substituting the original variable intervals also provides new limits of integration, maintaining the definite nature of the integral.

This substitution not only simplifies the form of the integrand but also makes it more convenient to employ further techniques, like trigonometric substitution.

Definite Integrals

Definite integrals are integral expressions bounded by specific numerical limits. Evaluating definite integrals provides the actual numerical value representing the area under the curve.
Once the indefinite integral \( \theta + \frac{1}{2} \sin 2\theta + C \) is found, evaluating this result between the new limits \( \sqrt{1/12} \) and \(1/2\) is the final step.

• Calculate the antiderivative at these limits, \( \theta|_{\sqrt{1/12}}^{1/2} \), by substituting \( \theta\) values that correspond to these \(u\) values.
• Trigonometric functions require converting expressions back to the original variable state to obtain the rightful numerical answer.
• The eventual result is a number which represents the area between the provided limits, \(\frac{1}{12}\) and \(\frac{1}{4}\) in the original integral context.

Definite integrals provide valuable insights, such as area, which is important in practical applications.