Problem 66 - Calculus Early Transcendentals

Question

For the following functions \(f\), find the antiderivative \(F\) that satisfies the given condition. $$f(\theta)=2 \sin 2 \theta-4 \cos 4 \theta ; F\left(\frac{\pi}{4}\right)=2$$

Step-by-Step Solution

Answer

Answer: The antiderivative of the function is \(F(\theta) =-\frac{1}{2}\cos(2\theta) -\sin(4\theta) + 2\).

Step 1: Integrate the trigonometric functions

We must first integrate \(f(\theta)\) with respect to \(\theta\). To do this, recall the antiderivative rules for sine and cosine functions: Integral of \(\sin(ax)\) is \(-\frac{1}{a}\cos(ax) + C\) Integral of \(\cos(ax)\) is \(\frac{1}{a}\sin(ax) + C\) Using these rules, we will integrate the given function: $$F(\theta) = \int(2\sin2\theta - 4\cos4\theta)d\theta$$ $$F(\theta) = -\frac{1}{2}\cos(2\theta) -\frac{4}{4}\sin(4\theta) + C$$ Simplify the result: $$F(\theta) = -\frac{1}{2}\cos(2\theta) -\sin(4\theta) + C$$

Step 2: Incorporate the given condition

The given condition is \(F\left(\frac{\pi}{4}\right)=2\). Plug in \(\frac{\pi}{4}\) into the formula for \(F(\theta)\) and set it equal to 2: $$2 = -\frac{1}{2}\cos(2\cdot \frac{\pi}{4}) - \sin(4\cdot \frac{\pi}{4}) + C$$ $$2 = -\frac{1}{2}\cos(\frac{\pi}{2}) - \sin(2\cdot \pi) + C$$

Step 3: Solve for the constant C

Using the trigonometric values of \(\sin(2\pi) = 0\) and \(\cos(\frac{\pi}{2}) = 0\), we can easily solve the equation: $$2 = -\frac{1}{2}(0) - 0 + C$$ $$2 = C$$

Step 4: Write the final antiderivative function

By solving for C and inserting it back into the \(F(\theta)\) equation, we have the final antiderivative function: $$F(\theta) =-\frac{1}{2}\cos(2\theta) -\sin(4\theta) + 2$$

Key Concepts

Trigonometric Functions and Their Antiderivatives

Trigonometric functions like sine and cosine are foundational in calculus, particularly for finding antiderivatives. In the context of the function \(f(\theta) = 2 \sin 2\theta - 4 \cos 4\theta \), understanding these functions' antiderivatives is crucial. The antiderivative of \(\sin(ax)\) is \(-\frac{1}{a}\cos(ax) + C\), and for \(\cos(ax)\), it is \(\frac{1}{a}\sin(ax) + C\). These formulas are derived from basic differentiation rules and are inverse processes.

When you encounter problems requiring integration of trigonometric functions, always recall these fundamental antiderivative forms. The constants \(a\) in these functions affects the outcome significantly, as they determine how the original function stretches or compresses on the graph.

• Integrals of sine functions result in cosine functions but with the opposite sign.
• Integrals of cosine functions result in sine functions and maintain their original sign.

Identifying these correctly allows you to solve antiderivative problems smoothly and efficiently.

Integration Techniques for Trigonometric Functions

Integration is the reverse process of differentiation. For trigonometric functions, familiarity with specific formulas is necessary to integrate successfully:

To solve \(F(\theta) = \int(2\sin2\theta - 4\cos4\theta) d\theta\), you apply the knowledge of antiderivatives:

• Integrating \(2\sin2\theta\), you multiply the coefficient by the antiderivative of \(\sin2\theta\), resulting in \(-\frac{1}{2}\cos2\theta\).
• Integrating \(-4\cos4\theta\), you apply \(-\frac{4}{4}\sin4\theta\) to obtain \(-\sin4\theta\).

Each term is processed independently using linearity of integration and the chain rule in reverse.

It's essential to also recognize the constant of integration \(C\), accounting for unknowns lost to differentiation. This constant is later determined using initial conditions or boundary values, which we'll discuss next.

Applying Initial Conditions to Solve for Constants

Initial conditions are necessary to determine the exact form of an antiderivative. Without them, the solution can be a family of functions differing by a constant. For this problem, the condition \(F\left(\frac{\pi}{4}\right) = 2\) is crucial.

To apply it, substitute \(\theta = \frac{\pi}{4}\) into the integrated function \(F(\theta)\) and solve for the constant \(C\).

• Calculate \(-\frac{1}{2}\cos(2 \cdot \frac{\pi}{4}) = -\frac{1}{2}\cos(\frac{\pi}{2})\), which results in 0 since \(\cos(\frac{\pi}{2}) = 0\).
• Likewise, \(-\sin(4 \cdot \frac{\pi}{4}) = -\sin(\pi)\) yields 0 as \(\sin(\pi) = 0\).

Set the result equal to the initial condition value: 2 = 0 + 0 + \(C\), solving this gives \(C = 2\).

Initial conditions provide the specific "y-intercept" for the antiderivative, making it unique to the problem, resulting in the final function \(F(\theta) = -\frac{1}{2} \cos(2\theta) -\sin(4\theta) + 2\). Using these steps ensures a precise solution tailored to any given conditions.