Question

For the following problems, use the substitutions \(\tan \left(\frac{x}{2}\right)=t, d x=\frac{2}{1+t^{2}} d t, \sin x=\frac{2 t}{1+t^{2}}\), and \(\cos x=\frac{1-t^{2}}{1+t^{2}}\). \(\int \frac{d x}{3-5 \sin x}\)

Short answer

The solution involves substituting \(\tan\left(\frac{x}{2}\right) = t\) and simplifying the integral steps.

Step-by-step solution

Make the Substitution

Start by substituting the given transformations into the integral. Since \(dx = \frac{2}{1+t^2} dt\) and \(\sin x = \frac{2t}{1+t^2}\), replace these into the integral: \[\int \frac{1}{3 - 5\sin x} \, dx = \int \frac{1}{3 - 5\left(\frac{2t}{1+t^2}\right)} \cdot \frac{2}{1+t^2} \, dt.\]

Simplify the Expression

Simplify the expression inside the integral. First, simplify the denominator:\[3 - \frac{10t}{1+t^2} = \frac{3(1+t^2) - 10t}{1+t^2} = \frac{3 + 3t^2 - 10t}{1+t^2}.\] So, our integral becomes:\[\int \frac{2}{(1+t^2)(3 + 3t^2 - 10t)} \, dt.\]

Simplify Further

Factor and simplify the quadratic expression in the denominator. Rewrite the expression:\[3 + 3t^2 - 10t = 3(t^2 - \frac{10}{3}t) + 3.\]To facilitate easier integration, we can write this in a factorable form, completing the square if necessary, or look for substitutions to factor the expression.

Solve the Integral

Once the factoring or simplification is complete, use partial fraction decomposition or another integration technique suitable for the form you have. Without explicit simplification, these steps involve expressing the simplified quadratic form in terms that allow direct integration.

Substitute Back

After finding the antiderivative in terms of \(t\), substitute back the original variable \(x\) using \(t = \tan\left(\frac{x}{2}\right)\). This will yield the solution to the integral in terms of \(x\).

Key concepts

Integral Calculus

Integral calculus deals with the concept of finding the total size or value from small units. It involves operations on integrals, where you are essentially summing up an infinite number of infinitesimally small areas to compute total area, volume, or other quantities. In this exercise, we are given an integral to solve: \[ \int \frac{dx}{3-5 \sin x} \]Here, integrating with respect to \(x\), we apply a clever substitution involving trigonometric identities to simplify and solve the expression.

• By substituting \(\tan\left(\frac{x}{2}\right)=t\), we change the variable and the form of the integral to better tackle its complexity.
• This substitution can make the trigonometric functions linear in terms of \(t\), easing the path for integration.

With these transformations, we reframe the problem into a form that is manageable with standard integration techniques.

Trigonometric Identities

Trigonometric identities are key for transforming functions and simplifying integrals. These identities relate the trigonometric functions to one another. For example, in our substitution, we use:

• \(\sin x = \frac{2t}{1+t^2}\)
• \(\cos x = \frac{1-t^2}{1+t^2}\)

These identities are helpful not just for substitution but also for simplifying complex trigonometric expressions into forms that are easier to integrate. These transformations often turn complex sinusoidal behaviors into rational expressions. This effectively unlocks methods of calculus that are more practicable with algebraic forms rather than transcendental forms.

Partial Fraction Decomposition

Partial fraction decomposition is a crucial technique used to simplify integrals into a sum of terms that are straightforward to integrate. Once we've simplified a complex expression using trigonometric substitution, we often end up with a rational function that can be approached with partial fractions. To apply partial fraction decomposition, follow these steps:

• Express the rational function as a sum of fractions that have simpler, polynomial denominators.
• Factor the denominator fully if possible.
• Use algebra to solve for constants that will make the decomposition possible.

With the integrand in the form of partial fractions, the terms are much simpler to integrate directly, as they often resemble basic integral forms like \(\frac{1}{t}\) or \(t^n\). This approach is widely applicable when dealing with rational functions in calculus.

Antiderivatives

An antiderivative is a function whose derivative is the original function. In integral calculus, finding the antiderivative is often the ultimate goal as it represents solving the indefinite integral. After transforming our integral with substitution, our task becomes finding the antiderivative of the resulting expression.

• The process involves recognizing standard forms and applying integral rules or techniques such as substitution or partial fraction decomposition.
• Once the antiderivative is found in terms of the new variable \(t\), we convert it back to the original variable \(x\).

This back-substitution step ensures the solution is in terms of the original context of the problem. The final solution provides the indefinite integral, which includes an arbitrary constant \(C\), reflecting the family of all possible antiderivative functions for the given derivative function.