Problem 55 - Calculus: One and Several Variables

Question

Evaluate by the Wallis formulas: (a) \(\int_{0}^{\pi / 2} \sin ^{7} x d x\). (b) \(\int_{0}^{\pi / 2} \cos ^{6} x d x\).

Step-by-Step Solution

Answer

Using Wallis' formulas, we find that: (a) \(\int_{0}^{\pi / 2} \sin^{7}(x) dx = \frac{15}{48} \pi\). (b) \(\int_{0}^{\pi / 2} \cos^{6}(x) dx = \frac{8}{15} \pi\).

Step 1: Use the Wallis formula for odd n

Since n is odd (n=7), we can use the Wallis formula for odd n and obtain the value of the integral as follows: \[\int_{0}^{\pi / 2} \sin^7(x) dx = \frac{1 \cdot 3 \cdot 5}{2\cdot 4 \cdot 6} \times \frac{\pi}{2}\]

Step 2: Simplify the expression

Now, we can simplify this expression by multiplying and dividing the numbers: \[\int_{0}^{\pi / 2} \sin^7(x) dx = \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \times \frac{\pi}{2} = \frac{15}{48} \pi\] So, the value of the integral \(\int_{0}^{\pi / 2} \sin^7(x) dx\) is \(\frac{15}{48} \pi\). #b) Evaluating the integral with n=6#

Step 1: Use the Wallis formula for even n

Since n is even (n=6), we can use the Wallis formula for even n and obtain the value of the integral as follows: \[\int_{0}^{\pi / 2} \cos^6(x) dx = \frac{2\cdot 4}{3\cdot 5} \times \frac{\pi}{2}\]

Step 2: Simplify the expression

Now, we can simplify this expression by multiplying and dividing the numbers: \[\int_{0}^{\pi / 2} \cos^6(x) dx = \frac{2 \cdot 4}{3 \cdot 5} \times \frac{\pi}{2} = \frac{8}{15} \pi\] So, the value of the integral \(\int_{0}^{\pi / 2} \cos^6(x) dx\) is \(\frac{8}{15} \pi\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, often conceptualized as the reverse process of differentiation. It allows us to find quantities like areas under curves, accumulated totals, and other concepts that involve summation over a continuum.

Specifically, when you see an integral sign (), think of it as a signal to sum up an infinite number of infinitesimally small quantities. For example, in the exercise, integration is used to find the area under the curve of ) over the interval ).

Understanding the Wallis formulas requires a solid grasp of integration. These formulas provide shortcuts for calculating integrals of sine and cosine functions raised to a power, which would otherwise involve complex and lengthy calculations. They're especially helpful when dealing with trigonometric integrals whose exponents are integers, simplifying the integration process considerably.

Trigonometric Integrals

Trigonometric integrals involve the integration of trigonometric functions, such as sine and cosine, and are a common occurrence in calculus problems. They can be quite challenging due to their oscillatory nature and periodicity.

The Wallis formulas are a powerful tool for evaluating these integrals when the exponents are whole numbers. For an integral with a sine function raised to an odd power, the formula uses the pattern of multiplying alternating odd and even numbers. Conversely, for a cosine function raised to an even power, the pattern involves multiplying even numbers.

In the given exercise, the evaluation of ) and ) is greatly simplified by the Wallis formulas. The key to using these formulas is recognizing when the power of the sine or cosine function is even or odd, which dictates which version of the Wallis formula to apply.

Calculus

Calculus is a branch of mathematics that studies how things change. It's divided into two main parts: differential calculus, which deals with rates of change and slopes of curves; and integral calculus, which deals with the accumulation of quantities and the areas under and between curves.

The problems provided involve integral calculus, applying formulas derived from the work of John Wallis, an influential mathematician. These formulas are a testament to the interconnectedness of different areas of calculus, as they draw upon sequences, products, and the mathematical constant , which bridges the gap between algebraic and geometric concepts.

Students learning calculus are encouraged to understand not only how to apply such formulas but also the underlying principles that make them valid - an important aspect of higher-level mathematical education. The process of solving these integrals also serves as a reminder of the importance of knowing when to apply specific techniques in calculus, such as recognizing symmetry or periodicity in functions, to simplify complex problems.