Problem 23 - Biocalculus Calculus for the Life Sciences

Question

First make a substitution and then use integration by parts to evaluate the integral. \(\int_{\sqrt{\pi / 2}}^{\sqrt{\pi}} \theta^{3} \cos \left(\theta^{2}\right) d \theta\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{1}{2} - \frac{\pi}{4}\).

Step 1: Make a Substitution

We'll begin by making the substitution \( u = \theta^2 \). Then, differentiate it to find \( du \):\[ du = 2\theta d\theta \]Solving for \( \theta d\theta \), we get:\[ \theta d\theta = \frac{1}{2} du \]The limits of integration also change. When \( \theta = \sqrt{\pi/2} \), \( u = (\sqrt{\pi/2})^2 = \pi/2 \). When \( \theta = \sqrt{\pi} \), \( u = (\sqrt{\pi})^2 = \pi \). Thus, the integral becomes:\[ \int_{\pi/2}^{\pi} \theta^2 \cos(u) \frac{1}{2} du \]

Step 2: Express \( \theta^2 \) in Terms of \( u \)

Since \( u = \theta^2 \), multiplying both sides by 1/2 gives us \( \frac{1}{2} u = \frac{1}{2} \theta^2 \). Therefore, \( \theta^2 = u \). The integral now simplifies to:\[ \frac{1}{2} \int_{\pi/2}^{\pi} u \cos(u) \ du \]

Step 3: Apply Integration by Parts

Integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Here, let \( v = \sin(u) \) and \( dv = \cos(u) \, du \) to simplify the integral. Let \( w = u \) and therefore \( dw = du \). Substituting these into the formula, we get:\[ \int u \, \cos(u) \, du = u\sin(u) - \int \sin(u) \, du \]

Step 4: Simplify the Integral

The integral of \( \sin(u) \) results in:\[ \int \sin(u) \ du = -\cos(u) \]Plug this result into our integration by parts expression:\[ \int u \, \cos(u) \, du = u\sin(u) + \cos(u) \]Therefore, our expression for the integral becomes:\[ \frac{1}{2}(u \sin(u) + \cos(u)) \]

Step 5: Evaluate the Definite Integral

Now evaluate the expression from \( \pi/2 \) to \( \pi \):\[ \frac{1}{2} \left[ (\pi\sin(\pi) + \cos(\pi)) - \left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right) \right] \]Simplify the expression: \( \sin(\pi) = 0 \), \( \cos(\pi) = -1 \), \( \sin(\pi/2) = 1 \), and \( \cos(\pi/2) = 0 \):\[ \frac{1}{2} (-1 - \frac{\pi}{2}) = -\frac{1}{2} - \frac{\pi}{4} \]

Step 6: Final Result

Thus, the evaluated definite integral is:\[ -\frac{1}{2} - \frac{\pi}{4} \].

Key Concepts

Integration by Parts

Integration by parts is a powerful technique for solving complex integrals. It is particularly useful when the integrand is the product of two functions. To understand this method, think of it as the reverse of the product rule for differentiation. The basic formula is:

• \( \int u \, dv = uv - \int v \, du \)

In this context, you typically choose:

• \( u \): a function that simplifies when differentiated,
• \( dv \): a function that integrates easily.

After choosing \( u \) and \( dv \), differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \). Next, substitute these into the integration by parts formula. In our example, you solve \( \int u \, \cos(u) \, du \) by setting \( u = u \) and \( dv = \cos(u) \, du \), then integrating \( dv \) to find \( v = \sin(u) \). The result is:

• \( u\sin(u) - \int \sin(u) \, du \)

Finally, calculate the remaining integral to complete the solution.

Definite Integrals

Definite integrals are used to calculate the total accumulation of a quantity, such as area under a curve, between two bounds. In this context, they differ from indefinite integrals because they include limits of integration. To evaluate a definite integral, use fundamental calculus principles:

• First, find the antiderivative.
• Then, apply the limits of integration.

In our worked example, the integral is initially \( \int_{\sqrt{\pi/2}}^{\sqrt{\pi}} \theta^3 \cos(\theta^2) \, d\theta \). After performing substitutions and integration by parts, evaluate the antiderivative as a definite integral from \( \pi/2 \) to \( \pi \). Use the result to determine the integral value over these bounds. The notation for this final step is:

• \( \left[ F(b) - F(a) \right] \)

where \( F \) is the antiderivative, \( b \) is the upper limit and \( a \) is the lower limit.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals that involve roots and are associated with trigonometric identities. This method involves substituting a trigonometric form for a variable expression, allowing the integral to be simplified using trigonometric identities.Here's how it works:

• Identify the parts of the integrand that resemble trigonometric identities.
• Choose a substitution based on known relationships, such as \( x = a\sin(\theta) \) for expressions involving \( \sqrt{a^2 - x^2} \).

In this exercise, the substitution was initially \( u = \theta^2 \), attempting to simplify the integral by changing the variable. This is slightly different from classical trigonometric substitution, but it highlights the versatility of substitution methods in calculus. The process leads to an integral that is easier to handle through methods like integration by parts, eventually simplifying to solve the problem. Always remember to adjust limits too, as the variable changes.