Problem 153
Question
\(\int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x\) is equal to (A) \(4 \log \frac{3}{4}\) (B) \(4 \log \frac{4}{3}\) (C) \(2 \log \frac{16}{9}\) (D) \(-\log \frac{81}{256}\)
Step-by-Step Solution
Verified Answer
Integral evaluates to 0; corresponding option not provided.
1Step 1: Analyze the Expression
We start by examining the expression under the square root: \[\sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2} - 2}\]By expanding it, we notice it is a common form when dealing with trigonometric identities resembling \( \cos(\theta) \) expanded.
2Step 2: Simplify the Expression with Trigonometric Identity
The expression \( \left(\frac{x+1}{x-1}\right)^{2} + \left(\frac{x-1}{x+1}\right)^{2} - 2 \) can be rewritten using the identity for \( \cos(\theta) \):\[onumber\left(\tan^2 \theta + \cot^2 \theta \right) - 2 = \sec^2 \theta - \csc^2 \theta\]This leads to a form of zero since:\[(\sec^2 \theta - \csc^2 \theta) - 2 = 0\]This is because the expressions are squares of inverses, leading to a value \( \sec \theta \csc \theta = 1 \). Thus the original simplifies to 0.
3Step 3: Conclude the Integral
Since the under-root expression is zero across the given range, the entire integral simplifies because \[\int_{-1 / 2}^{1 / 2} 0 \, dx = 0\].
4Step 4: Identify the Correct Option
The result 0 corresponds to an option that matches 0 in logarithmic terms, which is the situation where the argument of a logarithm is 1. Evaluate the options for which the logarithm of the ratio equals 0.
Key Concepts
Trigonometric IdentitiesIntegration TechniquesIntegral Simplification
Trigonometric Identities
Trigonometric identities are mathematical expressions that relate various trigonometric functions to one another. They allow us to manipulate and simplify expressions involving angles and their corresponding trigonometric ratios. In the context of definite integrals, trig identities play a pivotal role in simplifying complex expressions. For instance, the problem at hand involves simplifying an expression under a square root which appears similar to a trigonometric identity. The expression can be connected to the identity involving secant and cosecant functions, or closely related to \[(\tan^2 \theta + \cot^2 \theta) - 2 = \sec^2 \theta - \csc^2 \theta.\]By recognizing this form, we reduce the problem to evaluating simpler functions or, in some cases, it turns into a zero value, as seen here.
Integration Techniques
When faced with complex integrals, applying the right integration techniques is crucial for simplifying the process. In calculus, integration techniques might include substitution, partial fractions, or recognizing known forms that fit certain rules or theorems. In our scenario, the strategic application of a trigonometric identity simplifies the integral. By transforming the expression under the root to easier forms, integration becomes straightforward.
Key techniques often involve:
- Analyzing and rewriting the integral to fit known identities.
- Simplifying the integrand expression to a recognizable form (like a zero or a constant).
- Checking for symmetry or other properties that could simplify evaluation over an interval.
Integral Simplification
Integral simplification can immensely reduce the effort required to solve an integral. The simplification involves transforming the given integral into a form easier to evaluate. For definite integrals, simplification often includes the examination of the integrand inside the limits. Consider the integral mentioned:\[\int_{-1 / 2}^{1 / 2} \sqrt{ \left( \frac{x+1}{x-1} \right)^2 + \left( \frac{x-1}{x+1} \right)^2 -2 } \, dx \]Through simplification using trigonometric identities, this integral effectively becomes zero because the expression under the square root evaluates to zero for all x in the interval. Once the integrand is simplified to zero, the definite integral over any range will also resolve to zero, making it simple yet elegant in finding the solution.This process demonstrates that sometimes, complexities at first glance can often hide underlying simplicity, just a step of simplification away.
Other exercises in this chapter
Problem 151
\(\int_{0}^{\pi / 2} f(\sin 2 x) \sin x d x\) is equal to (A) \(\int_{0}^{\pi / 2} f(\sin 2 x) \cos x d x\) (B) \(\sqrt{2} \int_{0}^{\pi / 4} f(\cos 2 x) \cos x
View solution Problem 152
The absolute value of \(\int_{10}^{19} \frac{\sin x}{1+x^{8}} d x\) is (A) less than \(10^{-7}\) (B) more than \(10^{-7}\) (C) less than \(10^{-6}\) (D) more th
View solution Problem 154
If \(I_{n}=\int_{0}^{1} \frac{d x}{\left(1+x^{2}\right)^{n}} ; n \in N\), then (A) \(2 n I_{n+1}=2^{-n}-(2 n-1) I_{n}\) (B) \(2 n I_{n+1}=2^{-n}+(2 n-1) I_{n}\)
View solution Problem 155
Given \(f\) is an odd function and periodic with period \(2 .\) If \(f(x)\) is continuous \(\forall x\) and \(g(x)=\int_{0}^{x} f(t) d t\), then (A) \(g\) is an
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