Problem 62 - Calculus: Early Transcendentals

Question

Evaluate the definite integral. \( \displaystyle \int^{\pi/2}_{0} \cos x \sin(\sin x) \, dx \)

Step-by-Step Solution

Answer

The integral evaluates to \( 1 - \cos(1) \).

Step 1: Identify the Integral

The integral we need to evaluate is \( \int_{0}^{\pi/2} \cos x \sin(\sin x) \ dx \). We notice that it involves the product of \( \cos x \) and \( \sin(\sin x) \).

Step 2: Substitution Preparation

The integral suggests substitution might simplify it. A useful substitution here is letting \( u = \sin x \). This means \( du = \cos x \, dx \).

Step 3: Change Limits of Integration

With \( u = \sin x \), when \( x = 0 \), \( u = \sin 0 = 0 \); and when \( x = \pi/2 \), \( u = \sin(\pi/2) = 1 \). So, the limits change from \( 0 \) to \( 1 \).

Step 4: Rewrite the Integral

Using the substitution \( u = \sin x \) and \( du = \cos x \, dx \), we can rewrite the integral as follows: \( \int_{0}^{1} \sin u \, du \).

Step 5: Evaluate the New Integral

Now we evaluate \( \int_{0}^{1} \sin u \, du \). The antiderivative of \( \sin u \) is \( -\cos u \). So, we have \( -\cos u \bigg|_{0}^{1} \).

Step 6: Compute the Definite Integral

Substitute the limits into the antiderivative: \( -\cos(1) -(-\cos(0)) = -\cos(1) + 1 \). Thus, the value of the integral is \( 1 - \cos(1) \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in calculus for simplifying the evaluation of certain integrals. This method allows you to replace a difficult part of the integrand with a trigonometric function to make the integral easier to handle. In the case of the integral \( \int_{0}^{\pi/2} \cos x \sin(\sin x) \, dx \), substitution helps us move past the complexity stemming from the nested sine function. By setting \( u = \sin x \), we simplify the integral into a more manageable form.

• The derivative \( du \) is \( \cos x \, dx \), which conveniently matches part of our integrand.
• This turns our integral into \( \int \sin u \, du \), a much simpler expression to evaluate.

This method requires attention to detail, particularly with handling changed integration limits and differentiating the substitution correctly. Always remember to adjust the limits of integration according to the substitution so calculations remain accurate.

Evaluation of Integrals

Evaluating integrals, especially definite integrals, involves finding the antiderivatives and applying the fundamental theorem of calculus. In this problem, after substitution, the integral becomes \( \int_{0}^{1} \sin u \, du \).

• To solve \( \int \sin u \, du \), we use the antiderivative \( -\cos u \).
• Applying the limits 0 to 1 changes the integration into practical computation: evaluate \( -\cos u \) at these points.

Substitute the bounds back into the antiderivative to find: \[ -\cos(1) - (-\cos(0)) = -\cos(1) + 1 \] This calculation gives us the definite integral's value, illustrating how changing the problem's form, through substitution, aids in the evaluation process.

Calculus Problem Solving

Problem-solving in calculus often requires a step-by-step approach that relies on understanding and applying various techniques. In this exercise, we combined substitution and evaluation to solve an integral, emphasizing the importance of using these methods effectively.

• Identify any potential substitutions that can simplify the problem.
• Change variables carefully, updating integration limits to match the substitution.
• Find and apply the antiderivatives.
• Finally, compute the values using the limits of integration.

Being methodical in each step ensures precision and prevents errors. With practice, choosing the right techniques becomes second nature, transforming challenging calculus problems into more straightforward tasks. This example highlights how a mix of foundational calculus skills can solve complex integrals efficiently.