Problem 106 - Calculus Early Transcendentals

Question

If necessary, use two or more substitutions to find the following integrals. \(\int x \sin ^{4} x^{2} \cos x^{2} d x\left(\text {Hint}: \text { Begin with } u=x^{2},\) then use \right. \(v=\sin u .)\)

Step-by-Step Solution

Answer

Based on the given step-by-step solution, create a short answer question: Question: Use substitution to evaluate the integral \(\int x \sin ^{4} x^{2} \cos x^{2} d x\). Answer: \(\frac{1}{10}\sin^5 (x^2) + C\)

Step 1: First substitution: u = x^2

We begin by using the substitution \(u=x^2\). To replace the differential dx, we need to differentiate u with respect to x: $$\frac{d u}{d x} = \frac{d (x^{2})}{d x} = 2x.$$ Now, solve for dx: $$d x = \frac{d u}{2x}.$$

Step 2: Rewrite the integral with first substitution

Using \(u = x^2\) and \(dx = \frac{du}{2x}\), we rewrite the integral as: $$\int x \sin ^{4} x^{2} \cos x^{2} d x = \int x \sin ^{4} (u) \cos (u) \frac{d u}{2x} = \frac{1}{2} \int \sin ^{4} (u) \cos (u) d u.$$

Step 3: Second substitution: v = sin(u)

Now, let's use the second substitution, \(v = \sin u\). To replace \(du\), we need to differentiate v with respect to u: $$\frac{d v}{d u} = \frac{d (\sin u)}{d u} = \cos u.$$ Now, solve for \(du\): $$d u = \frac{d v}{\cos u}.$$

Step 4: Rewrite the integral with second substitution

Using \(v = \sin u\) and \(du = \frac{dv}{\cos u}\), we rewrite the integral as: $$\frac{1}{2} \int \sin ^{4} (u) \cos (u) d u = \frac{1}{2} \int v^{4} \cos (u) \frac{d v}{\cos u} = \frac{1}{2} \int v^{4} d v.$$

Step 5: Integrate the simplified integral

The integral is now in a simple form that can be directly integrated: $$\frac{1}{2} \int v^{4} d v = \frac{1}{2} \cdot \frac{1}{5}v^5 + C = \frac{1}{10}v^5 + C.$$

Step 6: Replace v and u with original variables

Now we need to replace \(v\) with \(u\) and then \(u\) with \(x\) to obtain the final answer. Recall that: $$v = \sin u \quad \text{and} \quad u = x^2.$$ So, replacing v with u, we get: $$\frac{1}{10}\sin^5 u + C = \frac{1}{10}\sin^5 (x^2) + C.$$

Key Concepts

Integral Calculus

Integral calculus is a fundamental component of mathematics that focuses on finding the accumulation of quantities and determining the area under a curve. When faced with functions that are not straightforward to integrate, such as products of trigonometric functions and polynomials, or compositions of functions, we require advanced techniques to solve them. The exercise given demonstrates a common scenario in integral calculus where direct integration is not possible, and substitution becomes a necessary strategy to simplify the integrand into a more manageable form for integration.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric functions such as sine, cosine, tangent, etc. These functions often appear in mathematical and real-world problems, typically oscillatory systems or wave patterns. In our exercise, the integrand consists of sine raised to a power and multiplied by cosine. To integrate such expressions, knowledge of trigonometric identities and integration formulas is essential. However, those integrals are typically not straightforward and may involve special techniques like the one in our problem: u-substitution.

U-Substitution Method

The u-substitution method is a powerful tool for integrating complex functions. It's a process that involves changing the original variable of integration to a new variable, making the integrand simpler to integrate. In the provided exercise, we use the substitution \( u = x^2 \) and \( v = sin(u) \), transforming the integral into one in terms of \( v \) that is easier to integrate. This method is analogous to finding the antiderivative of a composite function by using the chain rule in reverse.

Integration Techniques

Integration techniques encompass a variety of methods used to evaluate integrals that are not immediately integrable using basic antiderivatives. The exercise implements a two-step substitution process, which is a more advanced approach within the integration techniques. The choice of substitutions often requires insight into the function's structure and an understanding of which simplifications will lead to a solvable integral. Through skillful application of these techniques, complex integrals can be reduced to elementary forms and solved systematically.