Question

$$\text {Evaluate the following integrals.}$$ $$\int_{0}^{\pi / 2} \sqrt{1-\cos 2 x} d x$$

Short answer

Question: Evaluate the definite integral $$\int_{0}^{\pi / 2} \sqrt{1-\cos 2 x} d x$$. Solution: The value of the integral is $$\sqrt{2}$$.

Step-by-step solution

Simplify the integrand function

Recall the double-angle identity for cosine: $$\cos 2x = 1 - 2\sin^2x$$ Use this identity to simplify the integrand function: $$\sqrt{1 - \cos 2x} = \sqrt{1 - (1 - 2\sin^2x)} = \sqrt{2\sin^2x} = \sqrt{2} \cdot \sin x$$ So now our integral becomes: $$ \int_{0}^{\pi/2} (\sqrt{2} \cdot \sin x) dx$$

Find the antiderivative

Now we need to find the antiderivative of the integrand function. To do this, simply integrate the function with respect to x: $$ \int (\sqrt{2} \cdot \sin x) dx = \sqrt{2} \int \sin x dx$$ Remember that the integral of sin x is -cos x. Therefore, $$ \sqrt{2} \int \sin x dx = -\sqrt{2} \cos x + C $$

Evaluate the definite integral

Now we will evaluate the definite integral from 0 to π/2: $$ \int_{0}^{\pi/2} (\sqrt{2} \cdot \sin x) dx = -\sqrt{2}\cos x \Big|_0^{\pi/2}$$ To calculate the definite integral, we'll substitute the limits of integration into the antiderivative function and subtract: $$-\sqrt{2}\cos \left(\frac{\pi}{2}\right) - \left(-\sqrt{2}\cos \left(0\right)\right) = -\sqrt{2} \cdot 0 + \sqrt{2} \cdot 1 = \sqrt{2}$$ So, the value of the integral is $$\sqrt{2}$$: $$ \int_{0}^{\pi / 2} \sqrt{1-\cos 2 x} d x = \sqrt{2}$$

Key concepts

Trigonometric Integration

Trigonometric integration involves the process of integrating functions that contain trigonometric expressions. In our exercise, we start with the integral \( \int_{0}^{\pi / 2} \sqrt{1-\cos 2x} \, dx \). Integrating such functions often requires simplifying the expression to make the integration process easier.

Here, we utilize the double-angle identity for cosine, \( \cos 2x = 1 - 2\sin^2x \), to transform the expression under the square root. This simplification helps us convert a complex trigonometric integrand into a more manageable form.

Once we make this simplification, the integrand becomes \( \sqrt{2} \cdot \sin x \), which is straightforward to integrate.
Integrating trigonometric functions is a common task in calculus, requiring familiarity with basic trigonometric identities and integration formulas. Keeping these handy makes solving such integrals more approachable and less time-consuming.

Double-Angle Identity

The double-angle identities are powerful tools in trigonometry, used to simplify expressions involving trigonometric functions of double angles. For cosine, one such identity is \( \cos 2x = 1 - 2\sin^2x \). This formula allows us to express \( \cos 2x \) in terms of \( \sin x \), which can be simpler to work with in certain integrals.

In our example, applying this identity helps us break down the term inside the square root \( \sqrt{1 - \cos 2x} \). Simplifying using this identity results in \( \sqrt{2\sin^2x} \), which equals \( \sqrt{2} \cdot \sin x \).
Understanding and applying double-angle identities can significantly simplify the process of integration. It's useful to memorize these identities and recognize when they can transform an otherwise complex problem into a more straightforward one, as seen in this case.

Antiderivative Calculus

Antiderivative calculus involves finding a function whose derivative yields a given function. When working with definite integrals, finding the antiderivative is a crucial step to solve the integral. In our specific exercise, we needed to find the antiderivative of \( \sqrt{2} \cdot \sin x \).

The integral \( \int \sin x \, dx \) yields \(-\cos x + C\), where \(C\) is the constant of integration. Thus, integrating \( \sqrt{2} \cdot \sin x \) gives us \(-\sqrt{2} \cos x \).
For definite integrals, like \( \int_{0}^{\pi / 2} \sqrt{2} \cdot \sin x \, dx \), we evaluate the antiderivative at the bounds of the interval. In this example, substituting \( x = \pi/2 \) and \( x = 0 \) into \(-\sqrt{2} \cos x \) and calculating the difference gives the final result \( \sqrt{2} \).

Mastering antiderivative calculus is essential in determining the area under curves and solving real-world problems.