Question

$$\text {Evaluate the following integrals.}$$ $$\int_{0}^{\pi / 8} \sqrt{1-\cos 8 x} d x$$

Short answer

Answer: The exact value of the integral is $$\frac{1}{2} - \frac{\sqrt{2}}{4}$$

Step-by-step solution

Applying a trigonometric identity

Recall the double-angle identity for cosine function: $$\cos 2\theta = 1 - 2\sin^2\theta$$ Since we have $$\cos 8x$$ in the integral, we can rewrite the integral using $$4x = 2(\text{2x})$$. Hence, we have: $$\int_{0}^{\pi / 8} \sqrt{1-\cos 8 x} d x = \int_{0}^{\pi / 8} \sqrt{1-[1 - 2\sin^2(2x)]} d x$$

Simplify the integrand

Now we can rewrite the integral as: $$\int_{0}^{\pi / 8} \sqrt{2\sin^2(2x)} d x$$ And further simplify it to: $$\int_{0}^{\pi / 8} \sqrt{2}\sin(2x) d x$$ as the sine function has a positive value in the given limits.

Evaluate the integral

Now we can evaluate the integral using the antiderivative of the sine function: $$\int \sin(2x) dx = -\frac{1}{2}\cos(2x) + C$$ So, the integral becomes: $$\int_{0}^{\pi / 8} \sqrt{2} \sin(2x) d x = \sqrt{2} \left[-\frac{1}{2}\cos(2x)\right] \Big|_{\!0}^{\pi / 8}$$

Apply the limits of integration

Applying the limits of integration, we get: $$\sqrt{2} \left[-\frac{1}{2}\cos(2(\pi / 8)) -\left(-\frac{1}{2}\cos(2(0))\right)\right]$$ Simplifying further, we get: $$\sqrt{2}\left(\frac{1}{2}\cos(0) - \frac{1}{2}\cos(\pi / 4)\right) = \sqrt{2}\left(\frac{1}{2} - \frac{\sqrt{2}}{4}\right)$$ Multiplying by $$\sqrt{2}$$, we get: $$\frac{1}{2} - \frac{\sqrt{2}}{4}$$ Thus, the exact value of the given integral is: $$\int_{0}^{\pi / 8} \sqrt{1-\cos 8 x} d x = \frac{1}{2} - \frac{\sqrt{2}}{4}$$

Key concepts

Trigonometric Identities

Understanding trigonometric identities is crucial in simplifying complex trigonometric expressions, like the one in our example.
These identities are equations involving trigonometric functions that are true for every value of the occurring variables.

Here's a quick rundown of a pivotal identity used in our problem:

• Double-Angle Identity for Cosine: \(\cos(2\theta) = 1 - 2\sin^2(\theta)\).
  This is handy when transforming expressions like \(1 - \cos(8x)\) into something simpler to work with.

This identity lets us express cosine in terms of sine, simplifying the original expression to
\(\sqrt{1-[1 - 2\sin^2(2x)]} d x\), allowing us to proceed further with our calculations.

Trigonometric identities don't just stop at simplifying equations.
They are fundamental in measuring angles, relating geometric shapes, and solving numerous mathematical problems.
By mastering trigonometric identities, we unlock an essential toolset for tackling various mathematical challenges.

Definite Integrals

Definite integrals may appear daunting initially due to their complexity and distinct notation differences from indefinite integrals.
However, they offer a pathway to grasp both simple and intricate area calculations under curves.

The notation \(\int_{a}^{b} f(x)\, dx\) represents a definite integral, where

• \(\int\) symbolizes integration.
• \(a\) and \(b\) are limits of integration, defining the region on the x-axis.
• \(f(x)\) is the function being integrated (integrand).

In our example, the definite integral \(\int_{0}^{\pi / 8} \sqrt{2} \sin(2x) \, dx\) calculates the area under the function \(\sqrt{2} \sin(2x)\) from 0 to \(\pi / 8\).
Using definite integration, this exercise evaluates the exact area between the function and the x-axis within the specified limits.

Keep in mind that definite integrals offer a significant advantage in real-world applications like

• finding areas
• calculating total mass or charge
• solving problems on continual growth or decay

Mastery of definite integrals expands one's mathematical toolset considerably.

Antiderivatives

Antiderivatives serve as the foundational concept underlying the process of integration.
They are referred to as primitive functions or primitive integrals, typically denoted as \(F(x)\) for a function \(f(x)\).
When differentiated, they yield the original function.

Finding an antiderivative is often the first step in solving an integration problem.

An example relevant to our exercise is:

• \(\int \sin(2x) dx = -\frac{1}{2}\cos(2x) + C\)
  Here, \(-\frac{1}{2}\cos(2x)\) is an antiderivative of \(\sin(2x)\).

Understanding how to find antiderivatives allows one to transition between differential and integral functions easily.
In particular, the definite integral calculation involves substituting the upper and lower limits into the antiderivative and subtracting, which gives the total area calculated.

Grasping antiderivatives is crucial for solving various problems across calculus, including solving differential equations, interpreting physical phenomena, and in computational applications.