Problem 429 - Calculus Volume 1

Question

In the following exercises, compute each definite integral. $$\int_{0}^{1 / 2} \frac{\tan \left(\sin ^{-1} t\right)}{\sqrt{1-t^{2}}} d t$$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln \left( \frac{2}{\sqrt{3}} \right) \).

Step 1: Recognize the Substitution

Notice that the integral contains an expression involving \(\sin^{-1} t\). We can use the substitution \( u = \sin^{-1} t \), which implies \( t = \sin u \).

Step 2: Transform the Differential

Differentiate \( t = \sin u \) to find \( dt \). We have \( dt = \cos u \, du \). Therefore, substitute this into the integral so \( dt = \cos u \, du \).

Step 3: Change the Limits of Integration

When \( t = 0 \), \( u = \sin^{-1} 0 = 0 \). When \( t = \frac{1}{2} \), \( u = \sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6} \). This changes the limits of integration from 0 to \( \frac{1}{2} \) to 0 to \( \frac{\pi}{6} \).

Step 4: Substitute and Simplify the Integral

Substitute \( t = \sin u \) and \( dt = \cos u \, du \) into the integral. We have\[\int_{0}^{\frac{\pi}{6}} \frac{\tan u}{\sqrt{1-\sin^2 u}} \cos u \, du = \int_{0}^{\frac{\pi}{6}} \tan u \, du\]The expression \(\frac{1}{\sqrt{1-\sin^2 u}} = \frac{1}{\cos u}\), which simplifies the integral to \( \int \tan u \, du \).

Step 5: Integrate and Apply Limits

The integral of \( \tan u \) is \( -\ln|\cos u| \). Evaluate the indefinite integral and then substitute the limits:\[\left[ -\ln|\cos u| \right]_{0}^{\frac{\pi}{6}} = -\ln|\cos \frac{\pi}{6}| + \ln|\cos 0|\]

Step 6: Final Calculations

Calculate the values: \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \cos 0 = 1 \). Thus:\[-\ln \left(\frac{\sqrt{3}}{2}\right) + \ln(1) = -\ln \left(\frac{\sqrt{3}}{2}\right)\]Using the logarithm property, it simplifies to:\[\ln\left(\frac{2}{\sqrt{3}}\right)\]

Step 7: Final Answer

The definite integral evaluates to \( \ln \left( \frac{2}{\sqrt{3}} \right) \).

Key Concepts

Integration Techniques

Integration techniques are essential tools in calculus that help solve complex integrals. They provide a way to simplify and evaluate integrals that are not straightforward. There are several techniques available, including:

• Substitution: Replacing a complex function with a simpler one to make integration easier. This is particularly helpful when dealing with composite functions.
• Integration by Parts: Used when integrating the product of two functions, based on the product rule for differentiation.
• Partial Fractions: Decomposing a rational function into simpler fractions to facilitate easier integration.
• Trigonometric Substitution: A method used when the integrand involves quadratic expressions or trigonometric functions.

In the given exercise, we used the substitution method, where the substitution \( u = \sin^{-1}t \) simplifies the integrand. By transforming the integral into a simpler form, we could then integrate with respect to \( u \), illustrating the power of substitution as an integration technique.

Trigonometric Substitution

Trigonometric substitution is a technique in calculus for evaluating integrals that involve square roots of quadratic expressions. This method uses trigonometric identities to simplify these expressions, making integration more straightforward. Common substitutions include:

• \( x = a \sin \theta \)
• \( x = a \tan \theta \)
• \( x = a \sec \theta \)

In our exercise, we dealt with \( \sin^{-1}t \), leading us to use a substitution based on a trigonometric identity, specifically \( t = \sin u \). This is a direct application of trigonometric substitution, as it utilizes the identity \( 1 - \sin^2 u = \cos^2 u \) to simplify \( \sqrt{1-t^2} \). The simplification process is crucial, as it helps transform the integral into one involving standard trigonometric functions like \( \tan u \), which are easier to handle.

Limits of Integration

Limits of integration are the endpoints of the interval over which a definite integral is evaluated. They play a critical role in determining the value of the integral. When performing a substitution, it is vital to change the limits to match the new variable of integration. This ensures the integral remains consistent with the bounds of the original function.
In the context of our problem, after substituting \( u = \sin^{-1} t \), we had to determine the new limits for \( u \). Initially, when \( t = 0 \), \( u \) also equals 0, and when \( t = \frac{1}{2} \), \( u = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} \). These new limits are essential; they allow us to evaluate the integral in terms of \( u \), keeping the solution accurate and the result meaningful in the context of the definite integral. Transforming the limits with each substitution that you make ensures that the evaluation of the integral reflects the original problem accurately.