Problem 68 - Calculus Early Transcendentals

Question

Use integration by parts to derive the following formulas for real numbers \(a\) and \(b\) $$\begin{aligned} &\int e^{a x} \sin b x d x=\frac{e^{a x}(a \sin b x-b \cos b x)}{a^{2}+b^{2}}+C\\\ &\int e^{a x} \cos b x d x=\frac{e^{a x}(a \cos b x+b \sin b x)}{a^{2}+b^{2}}+C \end{aligned}$$

Step-by-Step Solution

Answer

Question: Derive the following formulas using integration by parts: 1. $$\int e^{ax}\sin bx\,dx = \frac{e^{ax} (a \sin bx - b\cos bx)}{a^2 + b^2} + C$$ 2. $$\int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2} + C$$ Answer: 1. To derive the formula for $$\int e^{ax}\sin bx\,dx$$, we applied integration by parts twice, choosing $$u=\sin bx$$ and $$dv=e^{ax}dx$$ in the first round, and $$u=\cos bx$$ and $$dv=\frac{b}{a}e^{ax}dx$$ in the second round. This led us to the formula $$\int e^{ax}\sin bx\,dx = \frac{e^{ax} (a \sin bx - b\cos bx)}{a^2 + b^2} + C$$. 2. To derive the formula for $$\int e^{ax}\cos bx\,dx$$, we applied integration by parts once, choosing $$u=\cos bx$$ and $$dv=e^{ax}dx$$, then substituting the result we derived in case 1 for the integral of $$e^{ax}\sin bx$$, leading us to the formula $$\int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2} + C$$.

Step 1: Identify u and dv

Choose \(u=\sin bx\) and \(dv=e^{ax}dx\). We'll now find \(du\) and \(v\).

Step 2: Compute du and v

To find \(du\), differentiate \(u\) with respect to \(x\) and multiply by \(dx\). To find \(v\), integrate \(dv\) with respect to \(x\): $$\begin{aligned} du = b\cos bx \,dx, \quad v = \int e^{ax}\,dx = \frac{e^{ax}}{a} \end{aligned}$$

Step 3: Apply integration by parts

Now apply the formula for integration by parts: $$\begin{aligned} \int e^{ax}\sin bx\,dx = \frac{e^{ax}\sin bx}{a} - \int \frac{e^{ax}}{a}b\cos bx\,dx \end{aligned}$$

Step 4: Apply integration by parts again

Now apply integration by parts again with \(u=\cos bx\) and \(dv=\frac{b}{a}e^{ax}dx\). Find \(du\) and \(v\): $$\begin{aligned} du=-b\sin bx\,dx, \quad v=\int \frac{b}{a}e^{ax}\,dx=\frac{e^{ax}}{a} \end{aligned}$$ Apply the formula for integration by parts again to the integral on the right: $$\int \frac{e^{ax}}{a}b\cos bx\,dx = \frac{e^{ax}\cos bx}{a} - \int \frac{e^{ax}}{a}(-b\sin bx)\,dx$$

Step 5: Substitute the result back

Now substitute the result back into our expression from Step 3: $$\begin{aligned} \int e^{ax}\sin bx\,dx = \frac{e^{ax}\sin bx}{a} - \left(\frac{e^{ax}\cos bx}{a} - \int \frac{e^{ax}}{a}(-b\sin bx)\,dx\right) \end{aligned}$$

Step 6: Simplify and rearrange

Combine terms and factor \(e^{ax}/(a^2+b^2)\) out of the integral: $$\begin{aligned} \int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} + C \end{aligned}$$ Case 2: Deriving the integral of \(e^{ax}\cos bx\)

Step 1: Identify u and dv

Choose \(u=\cos bx\) and \(dv=e^{ax}dx\).

Step 2: Compute du and v

Compute \(du\) and \(v\): $$\begin{aligned} du = -b\sin bx \,dx, \quad v = \int e^{ax}\,dx = \frac{e^{ax}}{a} \end{aligned}$$

Step 3: Apply integration by parts

Apply the formula for integration by parts: $$\begin{aligned} \int e^{ax}\cos bx\,dx = \frac{e^{ax}\cos bx}{a} - \int \frac{e^{ax}}{a}(-b\sin bx)\,dx \end{aligned}$$

Step 4: Simplify the integral

Simplify the integral on the right: $$\begin{aligned} \int e^{ax}\cos bx\,dx = \frac{e^{ax}\cos bx}{a} + \int \frac{e^{ax}}{a}b\sin bx\,dx \end{aligned}$$

Step 5: Substitute the result from Case 1

Now substitute the result from Case 1 for the integral on the right: $$\begin{aligned} \int e^{ax}\cos bx\,dx = \frac{e^{ax}\cos bx}{a} + \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} \end{aligned}$$

Step 6: Simplify and rearrange

Factor \(e^{ax}/(a^2+b^2)\) out of the expression and combine terms: $$\begin{aligned} \int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2+b^2} + C \end{aligned}$$

Key Concepts

Exponential Integrals

Exponential integrals often involve functions of the form \( e^{ax} \), where \( a \) is a constant. These integrals are pervasive in calculus due to their applications in fields like physics and engineering. The procedure for solving them can be made simpler by recognizing patterns and applying appropriate techniques like integration by parts.

• For exponential functions, the integral \( \int e^{ax} \, dx \) results in \( \frac{e^{ax}}{a} + C \), where \( C \) is the constant of integration.


• The exponential function \( e^{ax} \) does not vanish, and retains its exponential character after integration, which is significant in solving differential equations.


• They are particularly useful in deriving solutions that describe growth and decay processes, given their natural growth rate property.

Remember, the result is generally multiplied by the reciprocal of the coefficient of \( x \), in this case \( a \). Practicing the integration of exponential functions is key to mastering more complex scenarios, like combining them with trigonometric functions.

Trigonometric Integrals

Trigonometric integrals involve functions like \( \sin x \) and \( \cos x \).
These are essential components in integration by parts when coupled with exponential or polynomial functions. In the context of our problem, we are looking at exponential functions combined with sine and cosine.

• The integration by parts method typically breaks down the function into a product of simpler parts: a function \( u(x) \) and a differentials \( dv \). This approach is particularly useful when trigonometric functions are involved.


• The integrals \( \int \sin bx\, dx \) and \( \int \cos bx \ dx \) give rise to \( -\frac{1}{b}\cos bx + C \) and \( \frac{1}{b}\sin bx + C \) respectively.


• Combining these trigonometric identities with other functions helps to simplify more complex integrals.

These integrals can also emerge in periodic function analyses, harmonic motion, and signal processing cases.

Integration Techniques

Integration techniques refer to the various methods used to simplify and evaluate the integrals of complex expressions. The integration by parts technique is crucial when dealing with products of exponential and trigonometric functions.

• Integration by Parts: This is a technique where you choose parts of the integral according to \( u \) and \( dv \). It is based on the product rule for differentiation, and in formula form, it looks like \( \int u \, dv = uv - \int v \, du \).


• Choosing the right \( u \) and \( dv \) is crucial. Often, \( u \) is chosen as a function that becomes simpler when differentiated, and \( dv \) is a function that remains manageable when integrated.


• This method is particularly effective for integrating products of polynomial, exponential, and trigonometric functions, enabling us to break down complex integrals into simpler parts.

Beyond integration by parts, there are other techniques like substitution, partial fraction decomposition, and trigonometric identities, which also play roles in solving integrals depending on the functions involved.