Question

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

Short answer

\( \frac{1}{\sqrt{5}} \)

Step-by-step solution

Identify the substitution

Notice that the integrand contains the term \( \frac{1}{1+t^2} \), which resembles the derivative of \( \tan^{-1} t \), since \( \frac{d}{dt}(\tan^{-1} t) = \frac{1}{1+t^2} \). This hints that we can use substitution method.

Choose substitution variables

Let's set \( u = \tan^{-1} t \). Then, the differential \( du = \frac{1}{1+t^2} \, dt \), which matches part of our integrand.

Change limits of integration

Our original limits of integration are \( t = 0 \) to \( t = \frac{1}{2} \). When \( t = 0 \), \( u = \tan^{-1}(0) = 0 \). When \( t = \frac{1}{2} \), \( u = \tan^{-1}\left(\frac{1}{2}\right) \). Our new integral will go from \( u = 0 \) to \( u = \tan^{-1}\left(\frac{1}{2}\right) \).

Rewrite the integral

Substitute all terms to transform the integral in terms of \( u \):\[ \int_{0}^{\tan^{-1}\left(\frac{1}{2}\right)} \cos(u) \, du \]

Integrate and find the antiderivative

The integral of \( \cos(u) \) is \( \sin(u) \). Therefore, our integral becomes:\[ \sin(u) \bigg|_{0}^{\tan^{-1}\left(\frac{1}{2}\right)} \]

Evaluate the definite integral

Plug in the limits of integration:\[ \sin\left(\tan^{-1}\left(\frac{1}{2}\right)\right) - \sin(0) = \sin\left(\tan^{-1}\left(\frac{1}{2}\right)\right) \]Since \( \sin(0) = 0 \), we have:\[ \sin\left(\tan^{-1}\left(\frac{1}{2}\right)\right) \]

Simplify using trigonometric identities

Since \( \tan^{-1}(t) \) creates a right triangle with opposite side 1 and adjacent \( t \), we can use the identity \( \sin(\tan^{-1}(t)) = \frac{t}{\sqrt{1+t^2}} \). So, \( \sin\left(\tan^{-1}\left(\frac{1}{2}\right)\right) = \frac{\frac{1}{2}}{\sqrt{1+\left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{5}} \).

Key concepts

Substitution Method

The substitution method is a technique used to simplify the process of integration, especially when dealing with expressions that are difficult to integrate directly. The core idea is to change the variable of integration to make the integrand easier to handle.
In our exercise, we notice the presence of the expression \( \frac{1}{1+t^2} \), which resembles the derivative of \( \tan^{-1} t \). This observation suggests that a substitution could simplify the integral. We set \( u = \tan^{-1} t \), so the differential \( du = \frac{1}{1+t^2} \, dt \), matches a part of the integrand.
This substitution transforms the integral into a simpler form, in this case, \( \int \cos(u) \, du \), making it much easier to evaluate.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. These identities are extremely useful in simplifying expressions and solving trigonometric equations.
In our problem, once the substitution was made, we ended up with \( \sin(\tan^{-1}(t)) \). A common identity states that \( \sin(\tan^{-1}(t)) = \frac{t}{\sqrt{1+t^2}} \).
This identity helps us determine the definite value of the expression without directly calculating \( \sin \). By using geometric reasoning, based on right triangles, we identify the roles of the sides involved when \( \tan^{-1}(t) \) is considered as an angle of a right triangle.

Integration Limits

When you change variables in an integral, it's important to also change the limits of integration. The limits need to correspond to the new variable, after substitution.
In our exercise, we originally have \( t \) ranging from 0 to \( \frac{1}{2} \). With the substitution \( u = \tan^{-1} t \), we recalculate these limits. When \( t = 0 \), \( u = \tan^{-1}(0) = 0 \). When \( t = \frac{1}{2} \), \( u = \tan^{-1}\left(\frac{1}{2}\right) \).
The updated integral becomes \( \int_{0}^{\tan^{-1}(\frac{1}{2})} \cos(u) \, du \). Changing integration limits is crucial as it preserves the range over which the integral is evaluated, maintaining correctness of the result.

Inverse Trigonometric Functions

Inverse trigonometric functions play a pivotal role in calculus, especially when dealing with integrals involving trigonometric expressions. The most common are \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \). These functions return the angle whose trigonometric ratio is the given number.
In our example, the use of \( \tan^{-1} t \) is crucial. This function gives the angle whose tangent is \( t \).
Understanding the properties and applications of inverse trigonometric functions helps transform and evaluate integrals in cases where direct integration is complex. The properties of these functions allow us to relate angles directly to trigonometric values, such as when using the identity for \( \sin(\tan^{-1}(t)) \), completing the simplification process.