Problem 50 - Thomas Calculus

Question

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions in Exercises 49 and 50 . $$ \begin{array}{l}{\int \sqrt{1+\sin ^{2}(x-1)} \sin (x-1) \cos (x-1) d x} \\\ {\text { a. } u=x-1, \text { followed by } v=\sin u, \text { then by } w=1+v^{2}} \\ {\text { b. } u=\sin (x-1), \text { followed by } v=1+u^{2}} \\\ {\text { c. } u=1+\sin ^{2}(x-1)}\end{array} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{3} (1 + \sin^2(x-1))^{3/2} + C \).

Step 1: Choose the First Substitution

Let's start with substitution (a). Set \( u = x - 1 \). Then \( du = dx \). The integral becomes: \( \int \sqrt{1 + \sin^2(u)} \sin(u) \cos(u) \, du \).

Step 2: Apply the Second Substitution

Now use the second substitution from (a): \( v = \sin(u) \). Then \( dv = \cos(u) \, du \). Substitute to get: \( \int \sqrt{1 + v^2} \, v \, dv \).

Step 3: Implement Third Substitution

Use the third substitution from (a): \( w = 1 + v^2 \). This implies \( dw = 2v \, dv \) or \( v \, dv = \frac{1}{2} \, dw \). Substitute to obtain: \( \int \sqrt{w} \cdot \frac{1}{2} \, dw \).

Step 4: Simplify and Integrate

The integral simplifies to \( \frac{1}{2} \int w^{1/2} \, dw \). Integrating with respect to \( w \) gives \( \frac{1}{2} \cdot \frac{2}{3} w^{3/2} + C = \frac{1}{3} w^{3/2} + C \).

Step 5: Back Substitute

Replace \( w \) with \( 1 + v^2 \), giving \( \frac{1}{3} (1 + v^2)^{3/2} + C \). Substitute \( v = \sin(u) \), resulting in \( \frac{1}{3} (1 + \sin^2(u))^{3/2} + C \). Finally, back substitute \( u = x - 1 \) to get \( \frac{1}{3} (1 + \sin^2(x-1))^{3/2} + C \).

Key Concepts

calculus integration

Integration is a fundamental concept in calculus, often used to find areas, volumes, and other quantities that add up continuously. In calculus integration, the key idea is to sum an infinite number of infinitesimally small areas to reach a total area. This process is known as finding an antiderivative or indefinite integral. An antiderivative of a function gives us a new function, which, when differentiated, returns the original function. Generally, the integral of a function \( f(x) \) is expressed as \( \int f(x) \, dx \). The result is \( F(x) + C \), where \( F(x) \) is the antiderivative and \( C \) is an arbitrary constant, denoting the family of all antiderivatives.
When solving integrals, we employ various methods to simplify the process. One such method is substitution, which involves changing variables to make an integral easier to solve. This might involve replacing a composite part of the original function with a single variable. Integrals that appear complex at first glance sometimes become simple once we find a suitable substitution, aiding significantly in calculus problem solving.

techniques of integration

There are many techniques in integration designed to make solving integrals easier. One common technique is the **substitution method**, or **u-substitution**, which is analogous to the chain rule in differentiation. It's used when an integral contains a function and its derivative. The goal is to simplify the integral into a basic form by changing variables. The typical steps are:

• Identify a part of the integral that can be set as \( u \). This is usually an expression inside another function.
• Derive \( du \) from \( u \), allowing for substitution.
• Re-express the entire integral in terms of \( u \) and \( du \). Simplification often follows.
• Integrate in terms of \( u \).
• Back-substitute to return to the original variable.

For more complex integrals, we might have to perform multiple substitutions, as seen in the exercise solution's example with steps involving \( u, v, \) and \( w \). These layers of substitution help break down the problem gradually, turning a thorny integral into manageable stages.

calculus problem solving

Solving problems in calculus requires a strategic approach, particularly when dealing with integration. One mainstay strategy is **trial substitution**. This is where experimentation with different substitutions occurries, slowly simplifying the integral progressively. Each substitution should aim to strip away layers of complexity, revealing an easier form.
In the original exercise given to students, trying different sequences of substitutions might seem daunting, but it's about persistence and recognizing patterns. Clever substitutions convert intricate functions into simpler versions, aligning more closely with known integrals. Breaking down a calculus problem step by step is an essential skill.
To master calculus problem solving, one needs to become comfortable with various techniques and be adaptable. Integrals can be challenging, with some involving trigonometric identities or algebraic manipulation before they suit typical integration techniques. Familiarity with a wide array of methods ensures a more comprehensive toolkit, better equipping students to tackle any calculus problem they might encounter.