Problem 13 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

Berechnen sie a) \(\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x\) b) \(\int_{0}^{\frac{\pi}{2}} x \sin x \cos x d x\) Manchmal vereinfacht die partielle Integration das zu berechnende Integral, ohne direkt die gesuchte stammfunktion zu liefern. Fuhrt dann eine mehrfache Anwendung der partiellen Integration zum Ziel, so spricht man von einer "Reduktions forme\ell" oder "Rekursionsformet" Wir wollen dies an einem Beispiel erl?utern.

Step-by-Step Solution

Answer

a) \(\frac{\pi}{4}\) b) 0

Step 1: Identify method for part a

For part (a), we will use the trigonometric identity for double angles to simplify the integral: \(\cos^2 x = \frac{1 + \cos(2x)}{2}\). This allows us to rewrite the integral in a simpler form.

Step 2: Rewrite and integrate part a

Rewrite the integral using the identity, which gives \(\int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2x)}{2} \, dx\). This can be split into two separate integrals: \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx\). Integrate each part separately to get \(\frac{1}{2} \cdot \frac{\pi}{2} + 0 = \frac{\pi}{4}\).

Step 3: Choose method for part b

For part (b), the integrand is of the form \(x \sin x \cos x\). Notice we can simplify the product of trigonometric functions using the identity: \(2\sin x \cos x = \sin(2x)\). First, rewrite the integrand to \(\frac{x}{2} \sin(2x)\).

Step 4: Apply integration by parts to part b

Integration by parts is given by \(\int u \, dv = uv - \int v \, du\). Choose \(u = x\) and \(dv = \sin(2x)\, dx\). Then, \(du = dx\) and \(v = -\frac{1}{2}\cos(2x)\). Apply integration by parts: \(-\frac{x}{2}\cos(2x) \Big|_{0}^{\frac{\pi}{2}} + \frac{1}{2} \int (\cos(2x)) \, dx\).

Step 5: Simplify and integrate

Evaluate the first term: \(-\frac{\pi}{4} \cdot 0 + 0 = 0\). Now integrate the remaining part: \(\int \cos(2x) \, dx = \frac{1}{2}\sin(2x)\), leading to the final result of the definite integral \(0\).

Step 6: Final evaluation

For part a, the integral results in \(\frac{\pi}{4}\). For part b, the integral results in \(0\).

Key Concepts

Trigonometric Identities

In solving integrals involving trigonometric functions, using the right trigonometric identities can simplify the process significantly. For example, in part (a) of our problem, we used the identity for cosines, specifically the double angle formula: \( \cos^2 x = \frac{1 + \cos(2x)}{2} \). This identity was crucial because it turned a more complex square function into a simpler sum of terms.
Using trigonometric identities like this often allows you to rewrite complicated expressions in a more manageable form. Commonly used identities include the Pythagorean identities, angle sum and difference formulas, and half-angle identities.
When dealing with integration of trigonometric functions, recognizing these identities can help streamlining the integration process, making it easier and less prone to errors.

Integration Techniques

Integration techniques can vastly differ based on the type of integral you are evaluating. In our exercise, we employed two key techniques: rewriting using trigonometric identities and integration by parts.

• **Rewriting**: As shown in part (a), using trigonometric identities often simplifies the integration by changing the function into a more integrable form. This can turn integrals of squared functions into sums or differences that are easier to approach.
• **Integration by Parts**: This integral method comes from the product rule for differentiation. In part (b) of the problem, we used it by selecting parts of the integrated function as \(u\) and \(dv\). The formula \(\int u \, dv = uv - \int v \, du\) allowed us to simplify and solve more complex integrals by breaking them down into simpler parts.

Mastering these techniques requires practice and application across different problems. However, once learned, they become invaluable tools for solving definite and indefinite integrals.

Definite Integrals

Definite integrals, often seen with specified limits, provide the signed area under a curve between two points. For instance, in the exercise, we evaluated the definite integrals from \(0\) to \(\frac{\pi}{2}\).
These integrals have specific values, which means that unlike indefinite integrals, they result in real numbers. When solving a definite integral, it is crucial to assess the arrangement and limits correctly. The final evaluation must substitute these limits in to accurately calculate the area under the curve.
Calculation of definite integrals involves:

• Performing the integration process as usual to find the antiderivative.
• Substituting the upper and lower limits into this antiderivative.
• Subtracting the value at the lower limit from the value at the upper limit.

Understanding definite integrals also ties into real-world interpretations, such as finding distances, areas, and other quantities that rely on calculating the total accumulation of varying rates.