Chapter 5: Problem 30
Evaluate the integrals using Part 1 of the Fundamental Theorem of Calculus. $$ \int_{\pi / 6}^{\pi / 2}\left(x+\frac{2}{\sin ^{2} x}\right) d x $$
Short Answer
Expert verified
\( \frac{\pi^2}{9} + 2\sqrt{3} \)
Step by step solution
01
Identify the Function
Identify the function we need to integrate. In this case, it is \( f(x) = x + \frac{2}{\sin^2 x} \).
02
Find the Antiderivative
Find the antiderivative \( F(x) \) of the function \( f(x) \). For the given function \( f(x) = x + \frac{2}{\sin^2 x} \), the antiderivative is \( F(x) = \frac{x^2}{2} - 2 \cot(x) \).
03
Apply the Limits of Integration
Evaluate \( F(x) \) at the upper limit \( x = \frac{\pi}{2} \) and the lower limit \( x = \frac{\pi}{6} \).
04
Calculate \( F(\pi/2) \)
Substitute \( x = \frac{\pi}{2} \) into the antiderivative: \( F(\frac{\pi}{2}) = \frac{(\frac{\pi}{2})^2}{2} - 2 \cot(\frac{\pi}{2}) = \frac{\pi^2}{8} - 2 \times 0 = \frac{\pi^2}{8} \) since \( \cot(\frac{\pi}{2}) = 0 \).
05
Calculate \( F(\pi/6) \)
Substitute \( x = \frac{\pi}{6} \) into the antiderivative: \( F(\frac{\pi}{6}) = \frac{(\frac{\pi}{6})^2}{2} - 2 \cot(\frac{\pi}{6}) = \frac{\pi^2}{72} - 2 \times \sqrt{3} = \frac{\pi^2}{72} - 2\sqrt{3} \).
06
Subtract the Antiderivative Values
Calculate the definite integral by subtracting \( F(\pi/6) \) from \( F(\pi/2) \): \[ \int_{\pi/6}^{\pi/2} \left(x + \frac{2}{\sin^2 x}\right) dx = F(\pi/2) - F(\pi/6) = \frac{\pi^2}{8} - \left(\frac{\pi^2}{72} - 2\sqrt{3}\right) \].
07
Simplify the Expression
Simplify the expression: \[ \frac{\pi^2}{8} - \frac{\pi^2}{72} + 2\sqrt{3} = \frac{9\pi^2}{72} - \frac{\pi^2}{72} + 2\sqrt{3} = \frac{8\pi^2}{72} + 2\sqrt{3} = \frac{\pi^2}{9} + 2\sqrt{3} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
The antiderivative, often referred to as an indefinite integral, is a fundamental concept in calculus. It is essentially the reverse process of differentiation. When you find the antiderivative of a function, you are looking for a new function whose derivative equals the original function.
- For example, if you have a function \( f(x) \), finding its antiderivative means determining a function \( F(x) \) such that \( F'(x) = f(x) \).
- The antiderivative is not unique; it includes a "constant of integration" because functions like \( x^2 + C \) and \( x^2 + D \) (where \( C \) and \( D \) are constants) both have \( 2x \) as their derivative.
Definite Integral
The definite integral is a key concept you will often encounter in calculus. It allows you to evaluate the total accumulation of a quantity, such as area under a curve, between two specific points on the x-axis.
- The Fundamental Theorem of Calculus connects differentiation and integration, providing a powerful tool for evaluating definite integrals.
- The process involves taking the antiderivative of a function, evaluating it at the upper and lower limits, and finding the difference.
Trigonometric Functions
Trigonometric functions are not only central to trigonometry, but they also play a crucial role in calculus. Understanding these functions and their properties is essential for solving integration problems, especially those involving trigonometric identities.
- Functions like sine, cosine, and tangent, along with their counterparts cosecant, secant, and cotangent, appear frequently in calculus.
- In many integration problems, recognizing and manipulating these trigonometric functions is key to finding solutions.