Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Compute the following antiderivatives. If you solve the integral by a substitution, \(u(t)=\), then identify in writing \(u(t)\) and \(u^{\prime}(t)\). a. \(\int \sin ^{4}(t) \cos t d t\) b. \(\int t\left(1+t^{2}\right)^{3} d t\) c. \(\int \frac{1}{(z+1)^{3}} d z\) d. \(\int \frac{1}{4 t+1} d t\) e. \(\int t\left(1+t^{4}\right)^{3} d t\) \(f . \quad \int \frac{1}{3 z+1} d z\) \(g . \quad \int \frac{\sin t}{\cos t} d t\) h. \(\int(1+x)^{3} d x\) i. \(\int \frac{z}{z^{2}+1} d z\) \(j \cdot \int\left(1+t^{2}\right)^{3} t^{-1} d t\) k. \(\int e^{2+z} d z\) l. \(\int_{0}^{\pi} \sin (\pi+x) d x\) \(m . \int \sin (4 t) d t\) n. \(\int(\ln x) \frac{1}{x} d x\) o. \(\int e^{-z^{2}} z d z\)

Step-by-Step Solution

Answer

Apply appropriate substitutions and integration techniques for each integral to find the antiderivatives.

Step 1: Problem a - Simplify the Integral

Rewrite the integral \( \int \sin^4(t) \cos t \, dt \) as \( \int \sin^4(t) \cdot \cos(t) \, dt \). Let \( u(t) = \sin(t) \), then \( u'(t) = \cos(t) \). This suggests a substitution.

Step 2: Problem a - Substitute and Integrate

Substitute \( u = \sin(t) \) and \( du = \cos(t) \, dt \), changing the integral to \( \int u^4 \, du \). Integrate to get \( \frac{u^5}{5} + C \). Substituting back, the antiderivative is \( \frac{\sin^5(t)}{5} + C \).

Step 3: Problem b - Identify Substitution

Given \( \int t(1+t^2)^3 \, dt \), set \( u(t) = 1+t^2 \). Then, \( u'(t) = 2t \), and thus \( du = 2t \, dt \), or \( \frac{1}{2} du = t \, dt \).

Step 4: Problem b - Substitute and Integrate

The integral becomes \( \int (1/2) u^3 \, du \), which is \( \frac{1}{2} \cdot \frac{u^4}{4} + C = \frac{u^4}{8} + C \). Substituting back \( u = 1+t^2 \), we get \( \frac{(1+t^2)^4}{8} + C \).

Step 5: Problem c - Simplify the Integral

The integral \( \int \frac{1}{(z+1)^3} \, dz \) can be rewritten as \( \int (z+1)^{-3} \, dz \).

Step 6: Problem c - Integrate

Use the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \( n eq -1 \). Here, \( n = -3 \), so the integral is \( \frac{(z+1)^{-2}}{-2} + C = -\frac{1}{2(z+1)^2} + C \).

Step 7: Problem d - Integrate the Basic Form

The integral \( \int \frac{1}{4t+1} \, dt \) requires a simple substitution. Let \( u = 4t + 1 \), then \( du = 4 \, dt \) or \( dt = \frac{1}{4} \, du \).

Step 8: Problem d - Substitute and Integrate

The integral becomes \( \frac{1}{4} \int \frac{1}{u} \, du \), which is \( \frac{1}{4} \ln|u| + C = \frac{1}{4} \ln|4t+1| + C \).

Step 9: Problem e - Choose a Substitution

Given \( \int t(1+t^4)^3 \, dt \), set \( u(t) = 1+t^4 \). Then \( u'(t) = 4t^3 \), thus \( t \, dt = \frac{1}{4} t^3 \, du \). However, notice that derivative is not simple so treat substitution carefully.

Step 10: Problem e - Substitute and Integrate

The integral becomes not straightforward. Rearrange to get \( \int \frac{1}{4} u^3 \frac{1}{t} \, du \), requiring multiple steps or an adjustment in strategy. However, traditionally solve can be: assume derivative such as partial fractions after substitution.

Step 11: Problem f - Integrate the Basic Form

The integral \( \int \frac{1}{3z+1} \, dz \) requires substitution. Let \( u = 3z + 1 \), then \( du = 3 \, dz \) or \( dz = \frac{1}{3} \, du \).

Step 12: Problem f - Substitute and Integrate

Now, \( \int \frac{1}{u} \cdot \frac{1}{3} \, du \). Integrate to get \( \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln|3z+1| + C \).

Step 13: Problem g - Simplify and Integrate

The integral \( \int \frac{\sin t}{\cos t} \, dt \) can be written as \( \int \tan t \, dt \). The antiderivative of \( \tan t \) is \( -\ln|\cos t| + C \).

Step 14: Problem h - Expand and Integrate

Rewrite \( \int (1+x)^3 \, dx \) by expanding: \( 1 + 3x + 3x^2 + x^3 \). Integrate each term separately: \( x + \frac{3}{2}x^2 + \frac{3}{3}x^3 + \frac{1}{4}x^4 + C \).

Step 15: Problem i - Choose Substitution

For \( \int \frac{z}{z^2+1} \, dz \), let \( u = z^2+1 \). Then \( du = 2z \, dz \) or \( z \, dz = \frac{1}{2} du \).

Step 16: Problem i - Substitute and Integrate

Now integral \( \int \frac{1}{2} \frac{1}{u} \, du \) simplifies to \( \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|z^2+1| + C \).

Step 17: Problem j - Integrate using Logarithm Rule

Given \( \int \left(1+t^2\right)^3 t^{-1} \, dt \), recognize this as \( \int \left(1+t^2\right)^3 \cdot \frac{1}{t} \, dt \), try substitution type.

Step 18: Problem k - Rewrite the Integral

For \( \int e^{2+z} \, dz \), observe it similarly mixes exponential derivative forms. Rewrite it as \( \int e^z \cdot e^{2} \, dz = e^{2} \int e^z \, dz \).

Step 19: Problem k - Integrate

Integrate \( e^z \), obtaining \( e^{2} e^z + C \), giving \( e^{2+z} + C \).

Step 20: Problem l - Evaluate Definite Integral

\( \int_{0}^{\pi} \sin(\pi + x) \, dx \) is evaluated over a specific range. Use identity \( \sin(\pi + x) = -\sin x \), so integral is \( -\int_{0}^{\pi} \sin x \, dx \).

Step 21: Problem l - Evaluate

Integrate to get \( [-\cos(x)]_{0}^{\pi} = -[-1 - 1] = 2 \).

Step 22: Problem m - Integrate sine function

The integral \( \int \sin(4t) \, dt \) utilizes simple integration rule, \( \int \sin(at) \, dt = -\frac{1}{a} \cos(at) + C \).

Step 23: Problem m - Apply the Rule

Result is \( -\frac{1}{4} \cos(4t) + C \).

Step 24: Problem n - Identify Integration Strategy

For \( \int (\ln x) \cdot \frac{1}{x} \, dx \), notice derivative structure, indirectly suggests \( u = \ln x \), ni transformation possible, such that \( du = \frac{1}{x} \, dx \).

Step 25: Problem o - Determine Substitution

The integral \( \int e^{-z^2} z \, dz \) can use \( u = -z^2 \), hence \( du = -2z \, dz \). Adjust \( z \, dz \) to \( -\frac{1}{2} \, du \).

Step 26: Problem o - Substitute and Integrate

Integral becomes \( -\frac{1}{2} \int e^u \, du = -\frac{1}{2} e^u + C = -\frac{1}{2} e^{-z^2} + C \).

Key Concepts

Substitution Method

In calculus, the substitution method is a powerful technique used to simplify integrals, especially when they're difficult to solve directly. Imagine substitution as changing the perspective to make the problem easier to handle. It's akin to changing variables, which can simplify complex expressions.

To apply the substitution method, follow these basic steps:

• First, identify a part of the integral that can be substituted with a new variable, usually denoted by \( u \). This part is chosen because its derivative is also present in the integral.
• Next, compute the derivative \( u'(t) \) and express \( dt \) in terms of \( du \).
• Substitute \( u \) and \( du \) into the integral, transforming it into a simpler form.
• Finally, solve the integral with respect to \( u \), then substitute back to the original variable for the final solution.

A simple example is the integral \( \int \sin^4(t) \cos t \, dt \). Using \( u = \sin(t) \), we have \( du = \cos(t) \, dt \). Therefore, the integral simplifies to \( \int u^4 \, du \), which is easier to solve.

Integration Techniques

Integration techniques are diverse methods that help us find the antiderivatives of functions. Each technique is applicable under specific conditions, and familiarity with them allows for flexible problem-solving. Some of the most common techniques include:

• Basic Integration Rules: These are fundamental rules like the power rule, which states \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• Integration by Substitution: As discussed, substitution can simplify integrals by rearranging variables.
• Integration by Parts: This technique is derived from the product rule for derivatives. It's useful when dealing with products of functions.
• Partial Fraction Decomposition: This breaks down complex rational functions into simpler fractions, making them easier to integrate.

An example of applying basic rules is \( \int (z+1)^{-3} \, dz \). Here, we use the power rule resulting in \( -\frac{1}{2(z+1)^2} + C \), demonstrating straightforward application of integration techniques.

Definite Integrals

Definite integrals represent the accumulated quantity, such as area under a curve, over a specific interval. It's calculated from a lower to an upper limit, indicated by bounds \( a \) and \( b \), on the integral sign. The result is a numerical value, giving the net area under a curve from \( x = a \) to \( x = b \).One way to approach definite integrals is by using antiderivatives, which you might know better as the Fundamental Theorem of Calculus. After finding the antiderivative \( F(x) \) of the function \( f(x) \), we evaluate it at the upper and lower bounds:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a).\]For example, consider \( \int_{0}^{\pi} \sin(\pi + x) \, dx \). Using the identity \( \sin(\pi + x) = -\sin x \) simplifies the integral. You evaluate it using \( \cos(x) \) as the antiderivative. The definite integral becomes \( [-\cos(x)]_{0}^{\pi} = 2 \), representing the net area over that interval.