Problem 3 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int 2 x \cos (3 x-1) d x $$

Step-by-Step Solution

Answer

The integral is \( \frac{2}{3} x \sin(3x-1) + \frac{2}{9} \cos(3x-1) + C \).

Step 1: Choose u and dv

To use integration by parts, we need to choose parts of the integral to differentiate and integrate, respectively. Set \( u = 2x \) and \( dv = \cos(3x-1) \, dx \).

Step 2: Find du and v

Differentiate \( u \) and integrate \( dv \): \( du = 2 \, dx \) and to find \( v \), integrate \( dv \): \( v = \frac{1}{3} \sin(3x-1) \).

Step 3: Apply Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute \( u \), \( du \), \( v \), and \( dv \) into the formula:\[ \int 2x \cos(3x-1) \, dx = 2x \cdot \frac{1}{3} \sin(3x-1) - \int \frac{1}{3} \sin(3x-1) \, (2) \, dx \].

Step 4: Simplify and Compute Remaining Integral

Simplify the expression:\[ \frac{2}{3} x \sin(3x-1) - \frac{2}{3} \int \sin(3x-1) \, dx \].To solve \( \int \sin(3x-1) \, dx \), set \( w = 3x-1 \), then \( dw = 3 \, dx \), \( dx = \frac{1}{3} \, dw \).\[ \int \sin(w) \, \frac{1}{3} \, dw = -\frac{1}{3} \cos(w) + C \].Convert back to \( x \):\( -\frac{1}{3} \cos(3x-1) \).

Step 5: Final Integration

Replace back into the integral and simplify:\[ \frac{2}{3} x \sin(3x-1) + \frac{2}{9} \cos(3x-1) + C \].This is the antiderivative of the integral given.

Key Concepts

Definite and Indefinite Integrals

When dealing with integration, it's important to differentiate between definite and indefinite integrals. Indefinite integrals are expressions that represent a family of functions and include an arbitrary constant, typically denoted as \(+ C\). For instance, when integrating the function \(2x \cos(3x-1)\) by parts, we arrived at \(\frac{2}{3} x \sin(3x-1) + \frac{2}{9} \cos(3x-1) + C\). Here, the \(+ C\) signifies that the solution accounts for any constant that was part of the original function before differentiation.

• **Indefinite Integrals**: Represent a family of functions with an arbitrary constant.
• **Definite Integrals**: Evaluate to a specific number and include limits of integration. They calculate the area under the curve between two points.

In contrast, definite integrals not only involve the function but also specific limits, providing a numerical value as the result. If we were solving a definite integral, specific values for the limits would be plugged into the antiderivative, subtracting the lower limit from the upper limit for the final result. Recognizing these differences is crucial for correct evaluation.

Trigonometric Integration

Trigonometric functions like sine and cosine often appear in integrals, making trigonometric integration a vital skill to master. In our problem, the integral involved a cosine term, \(\cos(3x-1)\), over which we applied integration by parts.
Understanding how to integrate trigonometric functions necessitates a grasp of basic antiderivatives:

• **\( \int \sin(ax+b) \, dx = -\frac{1}{a} \cos(ax+b) + C \)**
• **\( \int \cos(ax+b) \, dx = \frac{1}{a} \sin(ax+b) + C \)**

Applying these rules allows for the integration of trigonometric components. In the exercise, integrating \(\cos(3x-1)\) yielded \(v = \frac{1}{3} \sin(3x-1)\). These calculations are grounded in understanding these formulas and substituting appropriately for different expressions, like changing variables when \(w = 3x-1\). This substitution simplifies integration and solves complex-looking integrals efficiently.

Antiderivatives

An antiderivative is the reverse operation of taking a derivative. If you think about differentiating as a process that finds the rate of change of a function, antiderivatives help us to go back to the original function from its derivative. In the context of integration, finding an antiderivative is solving an integral.
To express this concept further, consider the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]Here, finding \(v\) from \(dv\) through integration is essentially finding an antiderivative. In the example problem, \(v = \frac{1}{3}\sin(3x-1)\) is the antiderivative of \(dv = \cos(3x-1) \, dx\).

• The process may vary with complex functions, but the fundamental idea remains to track back to the original function.
• Integration by parts effectively breaks down more complicated integrals into simpler parts, establishing these antiderivatives clearly.

Always remember, the antiderivative must include the constant \(+C\) because differentiation nullifies constants; hence, integrating brings them back into consideration. Understanding antiderivatives is crucial not just for calculus problems, but also for applications in physics and engineering.