Problem 34 - Essential Calculus

Question

Use the specified substitution to find or evaluate the integral. $$ \begin{array}{l} \int_{0}^{1} \frac{d x}{2 \sqrt{3-x} \sqrt{x+1}} \\ u=\sqrt{x+1} \end{array} $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{\pi}{12}\).

Step 1: Apply the Substitution

Apply the provided substitution \(u=\sqrt{x+1}\), which implies \(x=u^2-1\). Also, derive the differential dx in terms of u. This yields \(dx=2u du\).

Step 2: Change the Limits of Integration

When \(x=0\), we get \(u=\sqrt{0+1}\), so the lower limit is 1. When \(x=1\), we get \(u=\sqrt{1+1}\), so the upper limit is \(\sqrt{2}\). Thus, the new limits of integration are 1 and \(\sqrt{2}\).

Step 3: Substitute Variables and Simplify the Integral

After applying the substitution, change the integral in terms of \(u\), \(\int_{1}^{\sqrt{2}} \frac{2u du}{2 \sqrt{3-(u^2-1)} u}\). Cancel out similar terms to simplify the integral to \(\int_{1}^{\sqrt{2}} \frac{du}{\sqrt{4-u^2}}\).

Step 4: Solve the Integral

The integral now represents the arc-sine form which is \( \int \frac{du}{\sqrt{a^2-u^2}} = \arcsin \left( \frac{u}{a} \right) + C\). Here \(a=2\). Thus, the integral becomes \(\arcsin \left(\frac{u}{2}\right)\) evaluated from 1 to \(\sqrt{2}\). Apply the limits to get \(\arcsin \left(\frac{\sqrt{2}}{2}\right) - \arcsin \left(\frac{1}{2}\right)\).

Step 5: Determine the Exact Value

\(\arcsin \left(\frac{\sqrt{2}}{2}\right)\) equals \(\frac{\pi}{4}\) and \(\arcsin \left(\frac{1}{2}\right)\) equals \(\frac{\pi}{6}\). Therefore, the result is \(\frac{\pi}{4} - \frac{\pi}{6}\) which simplifies to \(\frac{\pi}{12}\).

Key Concepts

Substitution Method

In integral calculus, the substitution method is a powerful technique used to simplify complicated integrals. When you have an integrand that appears too complex to evaluate directly, try to replace it with a simpler expression using a different variable. This is particularly helpful in trigonometric or radical expressions.

For example, in the given exercise, the substitution suggested is \(u = \sqrt{x+1}\). This choice helps because it simplifies the part of the integrand involving radicals. By making this substitution, you reshape the integral into a form that's easier to handle. The task is then to find \(x\) in terms of \(u\), giving \(x = u^2 - 1\), and to replace \(dx\) with \(2u \, du\), the derivative of \(x\) with respect to \(u\).

By doing this, you've effectively moved from an expression difficult to integrate, into one where the limits and the function inside are more tractable. Overall, substitution is about making choices that break down complexity into manageable parts.

Definite Integrals

Definite integrals, unlike indefinite integrals, are used to find the exact area under a curve between two bounds. They take limits of integration into account and provide a specific numerical result rather than a general expression with a constant of integration.

In this problem, we initially have the integral from \(x=0\) to \(x=1\). However, because of the substitution \(u=\sqrt{x+1}\), it's essential to adjust these limits to match the new variable. When substituting, the limit when \(x=0\) becomes \(u=1\) and when \(x=1\) becomes \(u=\sqrt{2}\).

Properly changing these limits is crucial as it allows the integration process to proceed correctly in the new variable framework. Failing to do so would result in incorrect evaluations and results. Therefore, ensuring limits correspond is an integral part of successful integration with substitution.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions like sine, cosine, tangent, and their reciprocals. These integrals often require clever substitutions or identities to simplify. In our exercise, after substitution and simplifying, we end up with an integral that looks like \( \int \frac{du}{\sqrt{4-u^2}} \).

This expression is a classic form of what leads to an inverse trigonometric function, specifically the arcsine function. The arcsine integral \( \int \frac{du}{\sqrt{a^2-u^2}} = \arcsin \left( \frac{u}{a} \right) + C\) is key, where \(a\) is a constant that is extracted from the form of the integral.

Once the integral is recognized as an arcsine function, it's straightforward to evaluate it within the given bounds. The exercise takes this through to completion, evaluating the arcsine at the transformed limits, resulting in precise trigonometric values, in this case, \( \frac{\pi}{4} - \frac{\pi}{6} \), which simplifies to \( \frac{\pi}{12} \). Understanding these transformations and recognizing these forms is fundamental to solving trigonometric integrals efficiently.