Question

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. \(\int_{0}^{\pi / 2} \sqrt{1-\cos (2 x)} d x\)

Short answer

The integral evaluates to \( \sqrt{2} \).

Step-by-step solution

Simplify the Integrand

The first step is to use the trigonometric identity for the cosine of double angles: \( \cos(2x) = 1 - 2\sin^2(x) \). Therefore, the expression \( \sqrt{1-\cos(2x)} \) can be simplified using \( 1-\cos(2x)= 2\sin^2(x) \). Hence, \( \sqrt{1-\cos(2x)} = \sqrt{2\sin^2(x)} = \sqrt{2} |\sin(x)| \). Since the integral is from 0 to \( \pi/2 \), where \( \sin(x) \) is non-negative, we can write \( \sqrt{2}\sin(x) \).

Rewrite the Integral

Now rewrite the given integral with the simplified form of the integrand: \[\int_{0}^{\pi/2} \sqrt{2} \sin(x) \, dx \] This can be simplified as \( \sqrt{2} \int_{0}^{\pi/2} \sin(x) \, dx \).

Evaluate the Trigonometric Integral

The antiderivative of \( \sin(x) \) is \( -\cos(x) \). Thus, computing the integral, we have:\[\sqrt{2} \left[ -\cos(x) \right]_{0}^{\pi/2}\] After evaluating at the bounds, this becomes: \[\sqrt{2} \left[-\cos\left(\frac{\pi}{2}\right) - (-\cos(0))\right]\]

Simplify the Result

Calculating the cosine values, we know \( \cos\left(\frac{\pi}{2}\right) = 0 \) and \( \cos(0) = 1 \). So,\[\sqrt{2} \left[ -0 + 1 \right] = \sqrt{2}\] Thus, the value of the integral is \( \sqrt{2} \).

Key concepts

Trigonometric Identities

When dealing with definite integrals, particularly those involving trigonometric functions, knowing trigonometric identities can greatly simplify the problem. In this exercise, one such identity is used to simplify the integrand. The identity for the cosine of double angles is: \( \cos(2x) = 1 - 2\sin^2(x) \). This identity transforms the expression \( \sqrt{1-\cos(2x)} \) into something more manageable. By substituting \( \cos(2x) \) with \( 1 - 2\sin^2(x) \), we find:

• \( 1 - \cos(2x) = 2\sin^2(x) \)
• \( \sqrt{1-\cos(2x)} = \sqrt{2\sin^2(x)} = \sqrt{2}\sin(x) \) since \( \sin(x) \) is positive between \( 0 \) and \( \pi/2 \)

Understanding and using such identities facilitates transforming the problem into a simpler integral, which is easier to evaluate.

Integral Calculus

Integral calculus is essential in computing areas under curves and in this case, evaluating definite integrals. A definite integral, like \( \int_{0}^{\pi / 2} \sqrt{2}\sin(x) \, dx \), is evaluated by calculating the antiderivative and then applying the fundamental theorem of calculus. The core part of the process involves:

• Recognizing the nature of our new integral based on the simplified expression \( \sqrt{2}\sin(x) \)
• Rewriting the integral in a way that takes advantage of known antiderivatives

This step focuses on the transition from a complex integral to a simplified form, allowing us to easily apply calculus techniques and solve it. Integral calculus, therefore, provides the language and methods for solving such problems by leveraging established mathematical rules.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse differentiation. Determining antiderivatives is crucial in solving definite integrals. In this exercise, the key function is \( \sin(x) \). The antiderivative of \( \sin(x) \) is \( -\cos(x) \). **Evaluating the Integral** involves:

• Using \( -\cos(x) \) to find the definite integral from 0 to \( \pi/2 \)
• Applying the fundamental theorem of calculus, which states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_a^b f(x) \, dx = F(b) - F(a) \)

Thus, the evaluation simplifies to substituting the bounds into \(-\cos(x)\) and computing this difference to find the area under the curve. This method allows us to calculate integrals efficiently once the antiderivative is known.