Problem 430 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The solution of the definite integral is \( \ln(2) \).

Step 1: Recognize Trigonometric Identity

Recall the trigonometric identity where \( \sin(\cos^{-1}(t)) = \sqrt{1-t^2} \). Using this identity will help us simplify the given function in the integral.

Step 2: Simplify the Integral

Observe that \( \tan(\cos^{-1}(t)) = \frac{\sin(\cos^{-1}(t))}{\cos(\cos^{-1}(t))} = \frac{\sqrt{1-t^2}}{t} \). Substituting this into the integral gives us \int_{1/4}^{1/2} \frac{\sqrt{1-t^2}}{t \cdot \sqrt{1-t^2}} dt = \int_{1/4}^{1/2} \frac{1}{t} dt\.

Step 3: Compute the Integral

The simplified integral is \int \frac{1}{t} dt\, which is a standard integral that results in \ln|t| + C\. The definite integral thus becomes \[ \left. \ln|t| \right|_{1/4}^{1/2} = \ln(1/2) - \ln(1/4). \]

Step 4: Evaluate the Definite Integral

Simplify the expression \ln(1/2) - \ln(1/4)\ in the previous step using the properties of logarithms: \ln(1/2) - \ln(1/4) = \ln\left(\frac{1/2}{1/4}\right) = \ln(2).\ The result of the definite integral is \( \ln(2) \).

Key Concepts

Trigonometric Identity

In the world of integration, recognizing and applying trigonometric identities can considerably simplify the problem at hand. In this exercise, the identity that comes to our rescue is:

• \( \sin(\cos^{-1}(t)) = \sqrt{1-t^2} \).

This relates an inverse trigonometric function, \( \cos^{-1}(t) \), back to a basic trigonometric function. By using this identity, we can express \( \tan(\cos^{-1}(t)) \) in an alternate form.

Since \( \tan(x) = \frac{\sin(x)}{\cos(x)} \), it follows that:

• \( \tan(\cos^{-1}(t)) = \frac{\sin(\cos^{-1}(t))}{\cos(\cos^{-1}(t))} = \frac{\sqrt{1-t^2}}{t} \).

This transformation is crucial as it allows us to represent \( \tan(\cos^{-1}(t)) \) in terms of \( t \), making the integration process much more straightforward.

Integration by Substitution

This is a powerful tool in calculus that helps in simplifying complex integrals. Essentially, the method requires substituting part of the integral with a different variable, which turns a complex expression into a simpler one.

In the problem we're tackling, once we applied the trigonometric identity to simplify \( \tan(\cos^{-1}(t)) \), our integral was simplified to:

• \( \int_{1/4}^{1/2} \frac{1}{t} dt \).

Notice how the troublesome square root disappears, and the integral's form is now familiar.

The approach consists of finding an appropriate substitution that simplifies the function. In our case, recognizing that the function \( \frac{1}{\sqrt{1-t^2}} \) becomes manageable by expressing \( \sin(\cos^{-1}(t)) \) in simpler terms was key.

Although not directly employing a new variable here, the principle of substitution lies in reformulating parts of the integrand, leading to easier integration.

Properties of Logarithms

The final part of the solution involves applying the properties of logarithms to evaluate the definite integral.

Once we simplified the integral to:

• \( \int \frac{1}{t} dt \)

we solved it to find:

• \( \ln|t| + C \).

To compute the definite integral from \( t = 1/4 \) to \( t = 1/2 \), we used:

• \([ \ln(1/2) - \ln(1/4) ] \).

By applying the property of logarithms that states \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), we further simplify the expression:

• \( \ln\left(\frac{1/2}{1/4}\right) = \ln(2) \).

Using these logarithmic properties helps to simplify and solve expressions quickly and accurately. Especially in definite integrals, where initial computations often require exact values, understanding these properties can be immensely beneficial.