Problem 44 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the definite integral. \(\int_{0}^{\pi / 2} \cos x \sin (\sin x) d x\)

Step-by-Step Solution

Answer

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

Step 1: Identify the integral structure

We need to evaluate \[\int_{0}^{\pi / 2} \cos x \sin (\sin x) \, dx.\]This is a definite integral from 0 to \(\pi/2\). The integrand involves a composition of functions with cosine and sine.

Step 2: Use substitution method

We notice that the function inside the sine is \(\sin x\). Let's choose \( u = \sin x \). Then, the differential of \( u \) is \( du = \cos x \, dx \), which matches the \( \cos x \, dx \) part of the integrand. This simplifies our integral significantly.

Step 3: Change limits of integration

With the substitution \( u = \sin x \), we need to change the limits of integration. When \(x = 0\), \(u = \sin 0 = 0\). When \(x = \pi/2\), \(u = \sin(\pi/2) = 1\). So the integral from 0 to \(\pi/2\) in terms of \(x\) corresponds to the integral from 0 to 1 in terms of \(u\).

Step 4: Rewrite the integral with new variable

Substituting \( u = \sin x \) and \( du = \cos x \, dx \) into the integral, we have:\[\int_{0}^{\pi/2} \cos x \sin(\sin x) \, dx = \int_{0}^{1} \sin(u) \, du.\]

Step 5: Integrate with respect to \(u\)

The integral \(\int \sin(u) \, du\) is straightforward. It evaluates to:\[-\cos(u) + C\quad (\text{where } C \text{ is the constant of integration for an indefinite integral}).\]For a definite integral, evaluate as:\[-\cos(u) \Big|_{0}^{1}.\]

Step 6: Evaluate the definite integral

Substitute the limits into \(-\cos(u)\) to get:\[-\cos(1) + \cos(0).\]Since \(\cos(0) = 1\), the definite integral becomes:\[1 - \cos(1).\]

Step 7: Write the final result

Thus, the value of the given definite integral is\[1 - \cos(1).\]

Key Concepts

Substitution Method

When solving integrals, especially definite integrals, the substitution method can simplify complex expressions, making integration much easier. This method is akin to changing variables in an equation to make the math more straightforward.
Here's how we use the substitution method:

• Identify a part of the integrand that can be simplified and make it a new variable, like choosing \( u = \sin x \).
• Find the differential of this new variable in terms of the old one. For example, if \( u = \sin x \), then \( du = \cos x \, dx \).
• Rewrite the original integral in terms of this new variable. This process often involves both the integrand and the limits of integration.

With substitution, you take a seemingly complex integral and turn it into one that's much easier to manage.

Integration Limits

When you substitute a variable in a definite integral, it's crucial to change the integration limits to reflect this new variable. The integration limits define the range over which you are calculating the area under the curve.
In the example provided:

• The original integration limits are 0 to \( \pi/2 \) for \( x \).
• We need to convert these limits for the new variable \( u \).
• Calculate the new limits based on the substitution: when \( x = 0 \), \( u = \sin 0 = 0 \); and when \( x = \pi/2 \), \( u = \sin(\pi/2) = 1 \).
• These new limits, 0 to 1, then replace the original in the integral, which now simplifies the calculation process.

This adjustment allows for a coherent transition between variables, ensuring the integrity of the definite integral.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are foundational in calculus due to their periodic properties and their occurrence in composition with other functions.
Understanding these basic functions can greatly help in solving integrals involving trigonometric expressions:

• Cosine \( \cos x \): Represents the adjacent over hypotenuse ratio in a right-angle triangle. It oscillates between -1 and 1 as its angle varies.
• Sine \( \sin x \): Represents the opposite over hypotenuse ratio. It also oscillates between -1 and 1, but with a phase shift compared to cosine.
• When integrated, these functions yield other trigonometric expressions (e.g., the integral of \( \cos x \) is \( \sin x + C \), where \( C \) is the constant of integration).

Grasping these basics can make recognizing patterns in integrands much more manageable.

Composition of Functions

The composition of functions occurs when the output of one function becomes the input for another. In integrals that include composition, identifying these layers becomes key to simplification.
For instance, consider the integral we are working with:

• The integrand, \( \cos x \sin(\sin x) \), is a combination involving both \( \cos x \) and a nested function \( \sin(\sin x) \).
• By using substitution, you peel back layers of complexity and solve the integral with greater ease.
• Recognize and match terms from different parts of the integrand with the differential, like \( du \) in our example.

This process of applying substitution to a composition leverages the natural hierarchy and dependency within functions, streamlining the integration process.