Question

In Exercises \(47-52,\) use the given trigonometric identity to set up a \(u\) -substitution and then evaluate the indefinite integral. $$\int \sin ^{3} 2 x d x, \quad \sin ^{2} 2 x=1-\cos ^{2} 2 x$$

Short answer

\(-\frac{1}{2} \cos 2x - \frac{1}{6} \cos^{3} 2x + C\)

Step-by-step solution

Separate Integral

Rewrite the original integral of \(\sin^{3} 2x\) as \(\int \sin^{2} 2x * \sin 2x dx\). This corresponds to the given identity \(\sin^{2} 2x = 1-\cos^{2} 2x\).

Set u-substitution

Set the substitution \(u = \cos 2x\). The derivative of this is \(du = -2\sin 2x dx\). As a result, we have that \(\sin 2x dx = -\frac{1}{2} du\).

Apply u-substitution

Apply the substitution to the integral, replacing both \(\sin^{2} 2x\) and \(\sin 2x dx\). This would make the integral become \(\int (1-u^{2})*-\frac{1}{2} du\). Simplifying gives us \(-\frac{1}{2}\int (1-u^{2}) du\).

Separate and Integrate

Separate the terms to simplify integration: \(-\frac{1}{2} \int du - \frac{1}{2} \int u^{2} du\). Now integrate to get \(-\frac{1}{2} u - \frac{1}{6} u^{3}\).

Back-substitute u

Replace \(u\) with the original expression, \(\cos 2x\), to get \(-\frac{1}{2} \cos 2x - \frac{1}{6} \cos^{3} 2x + C\). With \(C\) as the constant of integration.

Key concepts

U-Substitution

One of the fundamental techniques in calculus is u-substitution, which simplifies the process of finding the indefinite integral, also known as the antiderivative. Students often encounter complex functions that would be hard to integrate directly. That's when u-substitution becomes a powerful tool.

Imagine you have a function to integrate, and part of it looks like a derivative of another function within the integrand. You can make a 'u' substitution, where you let 'u' be that function. You then find the differential 'du', and replace the relevant parts of the original integral with 'u' and 'du'. This often leads to a simpler integral that is easier to solve. After integrating with respect to 'u', you substitute back to the original variable.

In the exercise provided, the application of u-substitution transforms the trigonometric integral into a polynomial one, which is straightforward to integrate. The beauty of u-substitution is that it can unravel even the most tangled integrands, making the integral much more manageable.

Trigonometric Integrals

Dealing with trigonometric integrals can be challenging, but by using identities and substitution, they become less daunting. These integrals involve trigonometric functions such as sine, cosine, tangent, and others. A common strategy for solving trigonometric integrals is the use of trigonometric identities, which allow us to rewrite the integrals in a more familiar form.

For example, in the solved problem, the identity \(\sin^{2}x = 1 - \cos^{2}x\) simplifies the integrand by reducing the powers of sine and converting them into a function of cosine. This particular strategy, combined with u-substitution, simplifies the original integral \(\int \sin^{3} 2x \, dx\) by reducing it to a polynomial expression in terms of \(u\), which is easier to integrate.

Mastering the use of trigonometric identities is crucial because it allows students to tackle a variety of integrals that would otherwise be difficult or even impossible to solve directly.

Integration Techniques

Integration techniques are a collection of tools at a student's disposal when facing the challenge of finding antiderivatives. Different integrals require different methods, and often, more than one method can be applied to the same integral, resulting in the same solution. Some of these techniques include u-substitution, integration by parts, partial fraction decomposition, and trigonometric substitution, to name a few.

It's essential to recognize which technique to apply to a given integral to solve it efficiently. As seen in the exercise, u-substitution was the optimal choice. Sometimes, an integral may need a combination of methods to arrive at a solution. Developing skills in these techniques allows for flexibility and adaptability when approaching a wide array of problems in calculus.

To enhance understanding, it's advisable to practice various integration techniques actively. This helps in identifying the best approach quickly and can be particularly useful in timed exam scenarios where students need to solve problems effectively and efficiently.