Problem 64 - Thomas Calculus

Question

Evaluate each integral in Exercises \(63-70\) by eliminating the square root. $$ \int_{0}^{\pi} \sqrt{1-\cos 2 x} d x $$

Step-by-Step Solution

Answer

The value of the integral is \(2\sqrt{2}\).

Step 1: Apply the Double Angle Formula

We start by simplifying the expression under the square root. The integrand involves a cosine of a double angle: \[ \sqrt{1 - \cos 2x} \]Using the trigonometric identity for cosine, we know that:\[ \cos 2x = 1 - 2\sin^2 x \]Substituting this identity into the integral, we have:\[ \sqrt{1 - (1 - 2\sin^2 x)} = \sqrt{2\sin^2 x} \] Which simplifies to:\[ \sqrt{2}\sin x \]

Step 2: Simplify the Integral

Replace the square root expression in the integrand using what we found in Step 1:\[ \int_{0}^{\pi} \sqrt{2}\sin x \, dx = \sqrt{2} \int_{0}^{\pi} \sin x \, dx \] This allows us to factor out the constant \(\sqrt{2}\), simplifying the integral to focus on integrating \(\sin x\).

Step 3: Integrate the Sine Function

Now, we need to compute the integral of \(\sin x\) from 0 to \(\pi\):\[ \int_{0}^{\pi} \sin x \, dx \]The antiderivative of \(\sin x\) is \(-\cos x\), so:\[ \left[ -\cos x \right]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = 1 + 1 = 2 \]

Step 4: Combine Results to Get the Final Answer

Using the result from integrating \(\sin x\), multiply by the constant \(\sqrt{2}\):\[ \sqrt{2} \cdot 2 = 2\sqrt{2} \]Thus, the evaluated integral is:\[ 2\sqrt{2} \]

Key Concepts

Double Angle Formula

In integral calculus, the double angle formula is a powerful tool, especially when dealing with trigonometric integrals. It allows us to rewrite trigonometric expressions involving double angles, simplifying complex trigonometric functions into manageable forms. For cosine, the double angle formula is:

• \( \cos 2x = 1 - 2\sin^2 x \)

With this formula, you can express \( \cos 2x \) in terms of \( \sin x \), which helps in simplifying integrals. In the solution provided for the exercise, we utilized the double angle formula to convert \( \cos 2x \) in the square root expression, \( \sqrt{1 - \cos 2x} \), into an expression involving \( \sin x \). This conversion is vital as it simplifies the integration process by reducing the complexity of the trigonometric expression. By substituting \( \cos 2x = 1 - 2\sin^2 x \), we transform our integral, making it easier to evaluate.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved parameters. They form the backbone of simplifying trigonometric expressions. Some crucial identities that are often used include:

• Pythagorean identities: such as \( \sin^2 x + \cos^2 x = 1 \)
• Angle sum and difference identities
• Double angle identities, like \( \cos 2x = 1 - 2\sin^2 x \) and \( \cos 2x = 2\cos^2 x - 1 \)

In the solution, we used the identity \( \cos 2x = 1 - 2\sin^2 x \) to simplify \( \sqrt{1 - \cos 2x} \) to \( \sqrt{2}\sin x \). Identifying and applying the correct trigonometric identity turns a complex looking integral into a more straightforward form, facilitating the integration process.

Integration Techniques

Integration techniques in calculus are methods used to find the antiderivative of functions, especially when dealing with complex or non-standard forms. Some popular integration techniques include:

• Substitution
• Integration by parts
• Trigonometric identities and substitution
• Partial fractions

In our specific solution, we made use of trigonometric identities as an integral technique. We transformed the integral \[ \int_{0}^{\pi} \sqrt{2}\sin x \, dx \]by first simplifying the expression using identities. Then, we factored out the constant \( \sqrt{2} \), allowing us to focus on integrating \( \sin x \). This is a straightforward example of how a complex trigonometric integral can be simplified initially by using identities, followed by basic integration procedures.

Antiderivative

The antiderivative, or the indefinite integral, of a function is a function whose derivative is the original function. It's like the reverse process of differentiation. Finding the antiderivative is essential to solving definite integrals and is expressed as:

• \( F(x) = \int f(x) \, dx \)

In our example, the antiderivative needed was for \( \sin x \). The antiderivative of \( \sin x \) is \( -\cos x \). Therefore, solving for the definite integral:\[ \int_{0}^{\pi} \sin x \, dx \]First involves finding and using this antiderivative, followed by evaluating it at the bounds 0 and \( \pi \). The solution then used the evaluation \[ \left[ -\cos x \right]_{0}^{\pi} = 1 + 1 = 2 \]Finally, multiplying the result by the constant \( \sqrt{2} \) from our earlier simplification step gave the final result of \[ 2\sqrt{2} \].Understanding how to find and use antiderivatives is fundamental in solving definite integrals in calculus.