Question

Perform the indicated integrations. $$ \int_{0}^{\pi / 2} \frac{\sin x}{16+\cos ^{2} x} d x $$

Short answer

The integral evaluates to \(-\frac{1}{\sqrt{17}} \ln \left( \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right)\)."

Step-by-step solution

Identify Appropriate Substitution

For the definite integral \( \int_{0}^{\pi / 2} \frac{\sin x}{16+\cos^{2} x} \, dx \), notice that the presence of \( \cos^2 x \) in the denominator suggests a trigonometric substitution or identities. We will use the identity \( \cos^2 x = 1 - \sin^2 x \) to rewrite the denominator as \( 16 + 1 - \sin^2 x = 17 - \sin^2 x \).

Substitute Variables

Let \( u = \sin x \). Then \( du = \cos x \ dx \), and since \( \sin x = u \), we have \( \cos^2 x = 1 - u^2 \). The limits of integration change from \( x = 0 \) to \( u = 0 \) and \( x = \frac{\pi}{2} \) to \( u = 1 \). The integral becomes\[\int_{0}^{1} \frac{u}{16 + (1-u^2)} \frac{du}{\sqrt{1-u^2}} \].

Simplify the Integral

Simplifying the integral, we have:\[\int_{0}^{1} \frac{u}{17 - u^2} \frac{du}{\sqrt{1-u^2}} \].This can be challenging with the current form, so let's consider another substitution or simplification technique.

Complete the Square in the Denominator

Consider rewriting the denominator \( 17 - u^2 \) as \( (\sqrt{17})^2 - u^2 \). This suggests using a trigonometric identity or integral table formula for evaluation.

Apply Trigonometric Identity

Using the identity for integrals of the form \( \int \frac{x}{a^2 - x^2} \, dx \), we recall:\[\int \frac{x}{a^2 - x^2} \, dx = \frac{-x}{a} \ln \left| \frac{a + x}{a - x} \right| + C\].In this case, \( a = \sqrt{17} \), so the integral evaluates to:\[\int \frac{u}{17 - u^2} \, du = -\frac{u}{\sqrt{17}} \ln \left| \frac{\sqrt{17} + u}{\sqrt{17} - u} \right| \Big|_{0}^{1}\].

Evaluate the Definite Integral

Substitute the limits into the evaluated integral:\(-\frac{1}{\sqrt{17}} \ln \left| \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right| - \left(-\frac{0}{\sqrt{17}} \ln \left| \frac{\sqrt{17} + 0}{\sqrt{17} - 0} \right|\right)\).Simplifying:\(-\frac{1}{\sqrt{17}} \ln \left( \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right)\).

Final Answer

The definite integral evaluates to:\(-\frac{1}{\sqrt{17}} \ln \left( \frac{\sqrt{17} + 1}{\sqrt{17} - 1} \right)\). This is the final answer.

Key concepts

Definite Integral

A definite integral is used to calculate the accumulated quantity or net area under the curve of a given function between two specified boundaries. In this exercise, we are tasked with integrating the function \( \frac{\sin x}{16 + \cos^2 x} \) over the interval from \( x = 0 \) to \( x = \frac{\pi}{2} \).
Definite integrals are essential for understanding accumulative processes, such as finding areas and solving physics or engineering problems.
In practice, to evaluate a definite integral, we first find an antiderivative to our function, then calculate the difference between its values at the upper and lower limits.Understanding the definite integral allows us to effectively manage changes in given quantities within bounded regions.

Trigonometric Identity

Trigonometric identities are equations that hold true for every angle value and are crucial tools in simplifying integrals that involve trigonometric functions. In our scenario, we encounter \( \cos^2 x \), embedded in the denominator.
By using the identity \( \cos^2 x = 1 - \sin^2 x \), we can convert expressions into more manageable forms. This makes the integration process more straightforward. Trigonometric identities help in transforming and simplifying integrals that otherwise appear complex.
It's often necessary to rewrite expressions using these identities for easier manipulation. This skill can be highly beneficial when dealing with other kinds of mathematical tasks, like solving equations or finding derivatives.

Integral Evaluation

Evaluating an integral, especially one that involves trigonometric functions, often requires simplifying the integrand and using known integral formulas. In this problem, we simplify our expression by recognizing it as a form that matches the integral type \( \int \frac{x}{a^2 - x^2} \, dx \).
By applying this standard integral formula, where \( a = \sqrt{17} \), we can evaluate the integral's indefinite form first: \( -\frac{x}{\sqrt{17}} \ln \left| \frac{\sqrt{17} + x}{\sqrt{17} - x} \right| + C \).
Integral evaluation is about handling complex expressions through formula recognition, substitution, and careful calculus.
This approach significantly eases evaluation once we familiarize ourselves with frequently occurring integral forms and results.

Substitution Method

The substitution method is a technique used in integration to transform a difficult integral into an easier one. Here, we chose \( u = \sin x \), which simplifies our definite integral to run over the interval from \( u = 0 \) to \( u = 1 \) as \( x \) ranges from \( 0 \) to \( \frac{\pi}{2} \).
With \( du = \cos x \, dx \), this substitution effectively reduces the expression's complexity, making it easier to manipulate. The substitution method is particularly useful when handling integrals involving trigonometric functions, algebraic expressions, or where direct integration isn't straightforward.
Mastering substitution provides a powerful tool for solving integrals by relating them to known forms or simpler functions.