Problem 55 - Calculus

Question

Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{0}^{\pi / 2} \sin x \sin (\cos x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify Substitution Variable

To use the substitution rule, identify a substitution that will simplify the integral. Let \( u = \cos x \). Then, \( du = -\sin x \, dx \).

Step 2: Adjust Differential and Limits of Integration

From the substitution \( u = \cos x \), we have \( du = -\sin x \, dx \), which implies \( -du = \sin x \, dx \). Substitute \( u \) and \( du \) into the integral and adjust the limits of integration. When \( x = 0 \), \( u = \cos 0 = 1 \). When \( x = \pi/2 \), \( u = \cos(\pi/2) = 0 \).

Step 3: Substitute and Re-write Integral

The integral \( \int_{0}^{\pi/2} \sin x \sin(\cos x) \ dx \) changes into \( \int_{1}^{0} \sin(u)(-du) \). The negative sign can be used to switch the limits, so the integral becomes \( \int_{0}^{1} \sin(u)\ du \).

Step 4: Integrate Using Basic Trigonometric Identity

Integrate \( \sin(u) \) with respect to \( u \), which is \( -\cos(u) \). Therefore, \( \int \sin(u) \ du = -\cos(u) + C \).

Step 5: Evaluate the Definite Integral

Evaluate the definite integral using the limits from 0 to 1: \( [-\cos(u)]_{0}^{1} = -\cos(1) + \cos(0) \).

Step 6: Compute Numerical Values

Compute the values: \( \cos(0) = 1 \) and \( \cos(1) \) is simply evaluated using a calculator for a numerical approximation, but we keep it exact as \( \cos(1) \). Thus, the result is \( 1 - \cos(1) \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is an insightful technique in integration that involves replacing variables in an integral with trigonometric functions. In our exercise, we start by identifying an appropriate substitution that simplifies the integration process. Here, we choose to set the substitution variable as \( u = \cos x \). This choice transforms the original integral into a simpler form. The derivative of this function is \( du = -\sin x \, dx \), which indicates how differentials can be adjusted in the substitution process. When conducting trigonometric substitution, it is vital to appropriately alter the differential in the integral, ensuring that all components of the original integrand are accounted for in terms of \( u \). This process helps in tackling integrals that contain products of sines, cosines, and other trigonometric expressions, allowing us to transform potentially complex integrals into more manageable forms.

Limits of Integration

When performing substitution in a definite integral, it is necessary to also adjust the limits of integration according to the new variable. In the original integral, the limits of integration are from \( x = 0 \) to \( x = \frac{\pi}{2} \). After substituting \( u = \cos x \), we need to find the new limits in terms of \( u \):

• When \( x = 0 \), \( u = \cos(0) = 1 \).
• When \( x = \frac{\pi}{2} \), \( u = \cos(\frac{\pi}{2}) = 0 \).

Thus, the new limits are from \( u = 1 \) to \( u = 0 \). However, the integration from a higher limit to a lower one requires adjusting the direction, which can be achieved by factoring a negative sign outside the integral. This adjustment results in flipping the limits, giving us a new integral from \( u = 0 \) to \( u = 1 \), making it easier to compute.

Calculating Definite Integrals

After substituting and adjusting the limits, we proceed to calculating the definite integral. The simplified integral is \( \int_{0}^{1} \sin(u) \, du \). This integration process uses basic trigonometric identities, where the integral of \( \sin(u) \) is \( -\cos(u) \). Computing the definite integral involves evaluating this antiderivative at the new limits.To solve \( \int_{0}^{1} \sin(u) \, du \), substitute the upper and lower bounds:

• Evaluate at 1: \( -\cos(1) \).
• Evaluate at 0: \( -\cos(0) \), which simplifies to -1.

Subtract the lower limit value from the upper limit value to get \(-\cos(1) - (-1) = 1 - \cos(1) \). Therefore, the original definite integral evaluates to \(1 - \cos(1)\), illustrating how through adept substitution and evaluation, we can determine the result of trigonometric integrals.