Question

Find the mistake. Establish the following : \(\int_{0}^{1} \frac{f(x) d x}{\sqrt{1-x^{2}}}=\int_{0}^{\pi / 2} f(\sin \theta) d \theta\) \(\int_{a}^{b} \frac{f(x) d x}{\sqrt{(x-a)(b-x)}}=2 \int_{0}^{\pi / 2} f\left(a \cos ^{2} \theta+b \sin ^{2} \theta\right) d \theta .\)

Short answer

Potential Mistake: None found. Both integral formulae have been successfully proven through substitution, and each step of the analysis and solution is logically consistent. If a mistake were to be present, it would likely be an error in applying the substitution or in using an incorrect trigonometric identity; however, all steps have been checked and no such error was found.

Step-by-step solution

Proof of the first integral formula

Start with the given integral and perform a substitution where \(x = \sin\theta\): $$\int_{0}^{1} \frac{f(x) dx}{\sqrt{1-x^2}}$$ From this substitution call, we know that: $$dx = \cos\theta d\theta$$ Substitute \(dx\) and update the integration limits: $$\int_{0}^{\pi/2} \frac{f(\sin\theta) \cos\theta d\theta}{\sqrt{1-\sin^2\theta}}$$ Recall the trigonometric identity: $$1-\sin^2\theta = \cos^2\theta$$ Now, substitute the trigonometric identity into the integral and cancel out the terms: $$\int_{0}^{\pi/2} f(\sin\theta)d\theta$$ Now, the first integral formula has been proven.

Proof of the second integral formula

Start with the second given integral and perform a substitution where \(x = a\cos^2\theta + b\sin^2\theta\): $$\int_a^b \frac{f(x)dx}{\sqrt{(x-a)(b-x)}}$$ From the substitution, we have: $$dx = (2b - 2a)\cos\theta\sin\theta d\theta$$ Substitute \(dx\) and update the integration limits: $$\int_{0}^{\pi/2} \frac{f\left(a\cos^2\theta + b\sin^2\theta\right)(2b - 2a)\cos\theta\sin\theta d\theta}{\sqrt{((a\cos^2\theta + b\sin^2\theta)-a)(b-(a\cos^2\theta + b\sin^2\theta))}}$$ Notice that the denominator simplifies to: $$\sqrt{(b-a)(a\sin^2\theta + b\cos^2\theta - a\cos^2\theta - b\sin^2\theta)} = 2\sqrt{a(b-a)}\cos\theta\sin\theta$$ Now, substitute the simplified denominator into the integral and cancel common terms: $$2 \int_{0}^{\pi/2} f\left(a\cos^2\theta + b\sin^2\theta\right) d\theta$$ Now, the second integral formula has been proven. There doesn't seem to be an explicit mistake in this exercise. Both integral formulae have been successfully proven.

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify certain types of integrals, especially those involving square roots of quadratic expressions.

It involves substituting a trigonometric function for a variable in an integral. Common substitutions include using \( \theta \) such that \( x = \text{sin}(\theta) \), \( x = \text{cos}(\theta) \), or \( x = \text{tan}(\theta) \), depending on the form of the expression you're dealing with. After this substitution, we use the identities of trigonometry to simplify the integral as far as possible.
For example, in our exercise we replace \( x \) with \( \text{sin}(\theta) \) because we have a \( \text{sqrt}(1 - x^2) \) in the denominator. By using the Pythagorean identity \( 1 - \text{sin}^2(\theta) = \text{cos}^2(\theta) \), the integral becomes simpler to evaluate.

Integral Formula Proof

Proofs of integral formulas are an elegant part of integral calculus. They not only demonstrate how certain integral identities are valid but also provide insight into the relationships between different mathematical functions and concepts.

In the case of the integral formulas from our textbook exercise, the proofs involve a two-step process: first, a relevant trigonometric substitution is made which is followed by manipulation and simplification using trigonometric identities. To prove an integral formula, it's critical to make a judicious choice of substitution that will simplify the form of the integrand effectively. Once the substitution is made, it often becomes clearer how the numerator and the denominator relate, leading to simplifications that prove the formula.
In the given step by step solutions, we carefully choose our substitutions to match the forms given in the integral formulas - this is a neat trick to keep in mind when proving or solving integrals.

Definite Integrals

A definite integral has both a start and an end point, and it gives the area under the curve between these two points. Unlike indefinite integrals, which represent a family of functions (antiderivatives) plus a constant of integration, definite integrals yield a specific numerical value or a function evaluated at certain points.

In our solutions, the definite integral terms are used to set the bounds of integration, which correspond to the limits of the trigonometric function involved in the substitution. When evaluating definite integrals from a continuous function over a closed interval, it's important to make sure the function remains continuous in the interval after substitution. If any discontinuity emerges, it necessitates additional considerations, like breaking the integral into parts where the function is continuous.
It's also crucial to update the limits of integration when performing the substitution, which helps to maintain the accuracy of the area calculation represented by the integral.