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Magazine Chapter 8: Problem 31
Evaluate the integrals. \(\int_{0}^{\pi / 2} \theta \sqrt{1-\cos 2 \theta} d \theta\)
Short Answer
Expert verified
The integral evaluates to \( \sqrt{2} \).
Step by step solution
01
Simplify the Expression Under the Square Root
Notice that \( \cos 2\theta \) is a double-angle formula, which can be rewritten as \( \cos 2\theta = 1 - 2\sin^2\theta \). Substitute this into the square root expression to get \( \sqrt{1 - (1 - 2\sin^2\theta)} = \sqrt{2\sin^2\theta} \). This simplifies to \( \sqrt{2} |\sin\theta| \). Given the limits from \(0\) to \(\pi/2\), where \(\sin\theta\) is positive, the expression further simplifies to \( \sqrt{2} \sin\theta \).
02
Rewrite the Integral
Substitute the simplified expression back into the integral: \[ \int_{0}^{\pi / 2} \theta \sqrt{2} \sin\theta \, d\theta. \] Factoring out the constant \(\sqrt{2}\), the integral becomes: \[ \sqrt{2} \int_{0}^{\pi / 2} \theta \sin\theta \, d\theta. \]
03
Apply Integration by Parts
For integration by parts, recall the formula: \( \int u \, dv = uv - \int v \, du \). Let \( u = \theta \) and \( dv = \sin\theta \, d\theta \). Then \( du = d\theta \) and \( v = -\cos\theta \). Applying integration by parts: \[ \int \theta \sin\theta \, d\theta = -\theta\cos\theta\Big|_0^{\pi/2} + \int \cos\theta \, d\theta \Big|_0^{\pi/2}. \]
04
Evaluate the First Part of Integration by Parts
Calculate \(-\theta\cos\theta\) from \(0\) to \(\pi/2\): \[ -\theta\cos\theta\Big|_0^{\pi/2} = -\left(\frac{\pi}{2}\cdot 0 - 0\cdot \cos 0\right) = 0. \] So the first term reduces to zero.
05
Evaluate the Second Part of Integration by Parts
Compute \(\int \cos\theta \, d\theta \) over \(0\) to \(\pi/2\): \[ \int \cos\theta \, d\theta = \sin\theta\Big|_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1. \]
06
Finalize the Integral
Combine the results from Steps 4 and 5 to find the value of the integral: \[ \int \theta \sin\theta \, d\theta = 1. \] Therefore, the whole integral becomes: \[ \sqrt{2} \cdot 1 = \sqrt{2}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a fundamental tool in calculus for finding the area under a curve between two specified points on the x-axis. In simpler terms, they help us determine the accumulated quantity, like an area, over an interval. In our original exercise, we are calculating a definite integral from 0 to \( \frac{\pi}{2} \).
When solving definite integrals, it's crucial to understand the limits of integration because they tell us the start and end points of the interval over which we are integrating. In our example, the limits are from 0 to \( \frac{\pi}{2} \), which already give us an idea that we are working with an area over a segment of the unit circle in the trigonometric plane.
When you evaluate the integral after applying any algebraic simplifications and calculus techniques, don't forget to plug these boundaries back into your final result to find the net area or value. This means after calculating through steps such as integration by parts, we substitute back and calculate the result for each term at the upper and lower limits, then find their difference. This is exactly how you arrive at the result of \( \sqrt{2} \) in the problem.
When solving definite integrals, it's crucial to understand the limits of integration because they tell us the start and end points of the interval over which we are integrating. In our example, the limits are from 0 to \( \frac{\pi}{2} \), which already give us an idea that we are working with an area over a segment of the unit circle in the trigonometric plane.
When you evaluate the integral after applying any algebraic simplifications and calculus techniques, don't forget to plug these boundaries back into your final result to find the net area or value. This means after calculating through steps such as integration by parts, we substitute back and calculate the result for each term at the upper and lower limits, then find their difference. This is exactly how you arrive at the result of \( \sqrt{2} \) in the problem.
Trigonometric Identities
Trigonometric identities allow us to transform and simplify expressions involving trigonometric functions. They are especially helpful in calculus because they can simplify complex expressions, making integration or differentiation easier.
In our exercise, we use the well-known identity for the double angle: \( \cos 2\theta = 1 - 2\sin^2\theta \). This identity helps us to simplify the original integral's expression \( \sqrt{1-\cos 2 \theta} \) into something more manageable: \( \sqrt{2\sin^2\theta} \), which simplifies further to \( \sqrt{2}\sin\theta \), assuming \( \sin\theta \) is positive because \( \theta \) is between 0 and \( \frac{\pi}{2} \).
Trigonometric identities are like tools in a toolbox that can transform complex problems into simpler ones, speeding up the process of solving integrals and other mathematical problems. It's essential to recognize and use these identities effectively for efficient problem-solving.
In our exercise, we use the well-known identity for the double angle: \( \cos 2\theta = 1 - 2\sin^2\theta \). This identity helps us to simplify the original integral's expression \( \sqrt{1-\cos 2 \theta} \) into something more manageable: \( \sqrt{2\sin^2\theta} \), which simplifies further to \( \sqrt{2}\sin\theta \), assuming \( \sin\theta \) is positive because \( \theta \) is between 0 and \( \frac{\pi}{2} \).
Trigonometric identities are like tools in a toolbox that can transform complex problems into simpler ones, speeding up the process of solving integrals and other mathematical problems. It's essential to recognize and use these identities effectively for efficient problem-solving.
Double-Angle Formulas
Double-angle formulas are specific types of trigonometric identities involving trigonometric functions of twice an angle. These formulas are very useful, as they help in transforming the functions into a more simplified form that can be integrated or differentiated more easily.
The double-angle formula for cosine is one of the most common: \( \cos 2\theta = \cos^2\theta - \sin^2\theta \) and the alternate forms are equally useful: \( \cos 2\theta = 1 - 2\sin^2\theta \) or \( \cos 2\theta = 2\cos^2\theta - 1 \). Each form might be more convenient depending on the context.
In our specific problem, the formula \( \cos 2\theta = 1 - 2\sin^2\theta \) was the key to simplifying \( \cos 2\theta \) into a function that we could easily take the square root of, eventually converting it into the integral form \( \int \theta \sqrt{2}\sin\theta \ d\theta \).
Knowing and understanding the double-angle formulas can greatly assist in tackling various trigonometric integration problems.
The double-angle formula for cosine is one of the most common: \( \cos 2\theta = \cos^2\theta - \sin^2\theta \) and the alternate forms are equally useful: \( \cos 2\theta = 1 - 2\sin^2\theta \) or \( \cos 2\theta = 2\cos^2\theta - 1 \). Each form might be more convenient depending on the context.
In our specific problem, the formula \( \cos 2\theta = 1 - 2\sin^2\theta \) was the key to simplifying \( \cos 2\theta \) into a function that we could easily take the square root of, eventually converting it into the integral form \( \int \theta \sqrt{2}\sin\theta \ d\theta \).
Knowing and understanding the double-angle formulas can greatly assist in tackling various trigonometric integration problems.
