Question

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

Short answer

\(\frac{1}{2} \ln(4/3)\)

Step-by-step solution

Identify the Integration Method

The given integral is \( \int_{0}^{1/2} \frac{\tan(\sin^{-1} t)}{\sqrt{1-t^2}} \, dt \). Notice that this expression involves \( \tan(\sin^{-1} t) \), which is a composite function. To solve this, we will convert it into a simpler form that is easier to integrate.

Simplify \( \tan(\sin^{-1} t) \)

We know that \( \sin^{-1} t = \theta \) implies \( \sin \theta = t \), and thus the right triangle relation \( \cos \theta = \sqrt{1-t^2} \). Therefore, \( \tan(\theta) = \frac{\sin \theta}{\cos \theta} = \frac{t}{\sqrt{1-t^2}} \). Substitute this back into the integral to get \( \int_{0}^{1/2} \frac{t}{1-t^2} \, dt \).

Integrate the Simplified Function

The integral now is \( \int_{0}^{1/2} \frac{t}{1-t^2} \, dt \). To solve this, use substitution: let \( u = 1-t^2 \), then \( du = -2t \, dt \), or \( t \, dt = -\frac{1}{2} du \). Transform the limits of integration; when \( t = 0, u = 1 \) and when \( t = 1/2, u = 3/4 \). Substituting these into the integral gives \( -\frac{1}{2} \int_{1}^{3/4} \frac{1}{u} \, du \).

Evaluate the Integral

We have \( -\frac{1}{2} \int_{1}^{3/4} \frac{1}{u} \, du \), which is \( -\frac{1}{2} [\ln|u|]_1^{3/4} \). Evaluating this, we get \( -\frac{1}{2} (\ln(3/4) - \ln(1)) \) = \( -\frac{1}{2} (\ln(3/4)) \). Simplifying further, we know \( \ln(1) = 0 \), so the answer is \( \frac{1}{2} \ln(4/3) \).

Final Answer

The calculated value of the definite integral is \( \frac{1}{2} \ln(4/3) \). This represents the area under the curve of the function from 0 to 1/2.

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify certain integrals by substituting trigonometric identities. It is particularly useful for handling expressions involving square roots and quadratic terms.
In this specific integral, we're dealing with the expression \( \tan(\sin^{-1} t) \). The substitution approach helps us to replace these composite functions with simpler trigonometric expressions.
When we recognize the inverse sine function \( \sin^{-1} t \) as an angle \( \theta \) in a right triangle, it leads to \( \sin \theta = t \). Therefore, other trigonometric identities can be used, like \( \cos \theta = \sqrt{1-t^2} \), which simplifies the integration process.
By further simplifying \( \tan \theta \) as \( \frac{t}{\sqrt{1-t^2}} \), we transform the integral into a more manageable form, allowing easier application of integration techniques.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \sin^{-1}(x) \), play a vital role in calculus, especially when dealing with integrals involving angles within right triangles.
These functions help us find the angle whose trigonometric function results in a given number. For example, \( \sin^{-1}(x) \) determines the angle whose sine is \( x \).
In the problem provided, the inverse sine function transforms the variable \( t \) into an angle \( \theta \). This transformation is fundamental because it allows us to use trigonometric identities to simplify and solve the integral.
When we express \( \sin^{-1} t = \theta \), we can derive other trigonometric values like \( \tan(\theta) \). This is crucial for integration since it translates complex expressions into simpler mathematical entities, reducing the difficulty of solving integrals.

Integration Techniques

Integration techniques are strategies employed to solve integrals, one of the fundamental operations in calculus used to find areas under curves.
In this exercise, after converting the integral's complex form using trigonometric identities, we utilize substitution, a cornerstone method in integration, to further simplify the problem.
Substitution involves changing variables to make an integral more tractable. Here, we set \( u = 1-t^2 \) so that the differential \( du \) transforms the integral\( \int \frac{t}{1-t^2} \, dt \) into \( -\frac{1}{2} \int \frac{1}{u} \, du \).
This change significantly aids in solving the integral, reducing it to a form that requires simple integration techniques. By applying basic integration rules, such as the natural logarithm rule, we evaluate the definite integral over the new limits, resulting in \( \frac{1}{2} \ln\left(\frac{4}{3}\right) \), capturing the solution in a straightforward way.