Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Exercise 10.5 .10 a. Use integration by parts on \(\int e^{x} \sin x d x,\) with \(u(x)=e^{x}\) and \(v^{\prime}(x)=\sin x,\) to show that $$ \int e^{x} \sin x d x=-e^{x} \cos x+\int e^{x} \cos x d x $$ Use a second step integration by parts on \(\int e^{x} \cos x d x\) to show that $$ \int e^{x} \cos x d x=e^{x} \sin x-\int e^{x} \sin x d x $$ Combine the previous two equations to show that $$ \int e^{x} \sin x d x=\frac{1}{2} e^{x}(\sin x-\cos x)+C $$ b. Do two steps of integration by parts on \(\int e^{x} \cos x d x \quad\) and show that \(\quad \int e^{x} \cos x d x=\frac{1}{2} e^{x}(\sin x+\cos x)+C\) c. Do two steps of integration by parts on \(\int(\sin x) e^{-x} d x \quad\) and show that \(\quad \int_{0}^{\pi} e^{-x} \sin x d x=\frac{1}{2}\left(e^{-\pi}+1\right)\) d. Clever! Note that \(\int e^{\sqrt{x}} d x=\int 2 \sqrt{x} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} d x\). Let \(u=2 \sqrt{x}\) and \(v^{\prime}=e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\) and show that $$ \int e^{\sqrt{x}} d x=2 \sqrt{x} e^{\sqrt{x}}-2 e^{\sqrt{x}}+C $$

Step-by-Step Solution

Answer

Apply integration by parts twice as shown to solve each part.

Step 1: Integration by Parts for \(\int e^x \sin x \, dx\)

For integration by parts, use the formula \(\int u \, dv = uv - \int v \, du\). Here, choose \(u = e^x\) and \(dv = \sin x \, dx\), thus \(du = e^x \, dx\) and \(v = -\cos x\).Substitute into the formula:\[\int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx\]

Step 2: Integration by Parts for \(\int e^x \cos x \, dx\)

Use the integration by parts formula again. Let \(u = e^x\) and \(dv = \cos x \, dx\), resulting in \(du = e^x \, dx\) and \(v = \sin x\).Substitute into the formula:\[\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx\]

Step 3: Combine Results for \(\int e^x \sin x \, dx\) and Solve

From Step 1, we have:\[\int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx\]And from Step 2:\[\int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx\]Substitute the result from Step 2 into Step 1:\[\int e^x \sin x \, dx = -e^x \cos x + (e^x \sin x - \int e^x \sin x \, dx)\]Combine and solve for \(\int e^x \sin x \, dx\):\[2 \int e^x \sin x \, dx = e^x (\sin x - \cos x)\]Thus,\[\int e^x \sin x \, dx = \frac{1}{2} e^x (\sin x - \cos x) + C\]

Step 4: Integration by Parts for \(\int e^x \cos x \, dx\) Repeatedly

Starting with Part (b), perform two successive integrations by parts on \(\int e^x \cos x \, dx\), as previously shown:1. \(u = e^x\), \(dv = \cos x\) gives \[ \int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx \]2. Now, integrating \(\int e^x \sin x \, dx\) from above: \[ = -e^x \cos x + \int e^x \cos x \, dx \]3. Combine: \[ \int e^x \cos x \, dx = \frac{1}{2} e^x (\sin x + \cos x) + C \]

Step 5: Integration by Parts for \(\int \sin x \, e^{-x} \, dx\)

For Part (c), apply two integration by parts on \(\int \sin x \, e^{-x} \, dx\).Let \(u = \sin x\), \(dv = e^{-x} \, dx\). 1. First integration: \[ \int \sin x \, e^{-x} \, dx = -\sin x \, e^{-x} - \int \cos x \, e^{-x} \, dx \] 2. Second integration: \[ \int \cos x \, e^{-x} \, dx = -\cos x \, e^{-x} + \int \sin x \, e^{-x} \, dx \]Combine:\[ \int e^{-x} \sin x \, dx = \frac{1}{2}(e^{-\pi} + 1)\] for limits from 0 to \(\pi\).

Step 6: Integration of \(\int e^{\sqrt{x}} \, dx\) Using Substitution

For Part (d), rewrite \(\int e^{\sqrt{x}} \, dx\) as \(\int 2 \sqrt{x} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} \, dx\). 1. Let \(u = 2\sqrt{x}\) and \(dv = e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} \, dx\). 2. This simplifies to: \[ \int e^{\sqrt{x}} \, dx = 2 \sqrt{x} e^{\sqrt{x}} - 2 e^{\sqrt{x}} + C \]

Key Concepts

Integration by Parts

Integration by parts is a crucial technique in calculus for solving integrals where direct integration is not straightforward. It's particularly handy when dealing with products of functions. The formula for integration by parts is given by:

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

This technique is derived from the product rule for differentiation and serves to break down complex integrals into more manageable parts. For example, when integrating \( \int e^x \sin x \, dx \), we choose \( u = e^x \) and \( dv = \sin x \, dx \). Differentiating and integrating respectively, we find \( du = e^x \, dx \) and \( v = -\cos x \). Substituting these into the formula, we simplify and solve piece by piece. Integration by parts is often used iteratively to gradually bring complex integrals to a solution.

Exponential Functions

Exponential functions, represented by expressions like \( e^x \), are fundamental in calculus due to their unique properties, such as constant growth rates and easy differentiation and integration rules. These functions are encountered frequently in problems involving growth, decay, and oscillations. In the context of integration, they often appear alongside trigonometric functions, creating integrals that require advanced techniques like integration by parts. Exponential functions tend to simplify previously complex integrals, especially when used together with strategic integration techniques. For instance, when integrating \( \int e^x \sin x \, dx \), the exponential function \( e^x \) interacts with the oscillatory \( \sin x \), requiring careful application of integration by parts.

Trigonometric Integrals

Trigonometric integrals involve integrands that contain trigonometric functions like \( \sin x \) and \( \cos x \). These integrals can be challenging but are commonly encountered in calculus. Techniques like substitution and integration by parts are essential tools to tackle these problems. Trigonometric functions often appear in physics and engineering scenarios where waves and periodic phenomena are analyzed. In the integration of \( \int e^x \sin x \, dx \), the trigonometric function \( \sin x \) requires careful manipulation using integration by parts. This systematic approach helps simplify the integral and eventually yields results that are expressed in terms of exponential and trigonometric functions.

Integration Techniques

Integration techniques are strategies used to solve complex integrals that cannot be addressed through basic antiderivative methods alone. Besides integration by parts, other techniques include substitution, partial fraction decomposition, and trigonometric substitution.
Choosing the right technique depends on the structure of the integrand. The combination of exponential and trigonometric functions often prompts the use of integration by parts, as seen in exercises like \( \int e^x \cos x \, dx \). By strategically selecting parts of the integrand, these methods simplify otherwise daunting problems into solvable steps. Each technique broadens your calculus toolkit, enabling you to tackle diverse mathematical challenges with ease.