Question

Prove that \(\int_{a}^{b} \frac{d x}{\sqrt{\\{(x-a)(b-x)\\}}}=\pi\), \(\int_{a}^{b} \frac{x d x}{\sqrt{\\{(x-a)(b-x)\\}}}=\frac{1}{2} \pi(a+b)\) (i) by means of the substitution \(\mathrm{x}=\mathrm{a}+(\mathrm{b}-\mathrm{a}) \mathrm{t}^{2}\), (ii) bymeans of the substitution \((\mathrm{b}-\mathrm{x})(\mathrm{x}-\mathrm{a})=\mathrm{t}\), and (iii) by means of the substitution \(x=a \cos ^{2} t\) \(+b \sin ^{2} t\)

Short answer

Based on the step-by-step solution, the short answer is as follows: Using three different substitution methods, we evaluated the given integrals and proved that they are equal to the given expressions. With substitutions \(x = a + (b-a)t^2\), \((b-x)(x-a)=t\), and \(x = a\cos^2t + b\sin^2t\), we found that the first integral evaluates to \(\pi\), the second integral evaluates to 0, and the third integral evaluates to \(\frac{1}{2}\pi (a+b)\).

Step-by-step solution

Substitute\(x = a + (b-a)t^2\) in the first integral

For the first integral, replace \(x\) with our substitution: \(x = a + (b-a)t^2\). Then, find \(dx = 2t(b-a)dt\), and plug these values into the integral. We also need to change the limits of integration according to the new variable \(t\). When \(x=a\), \(t=0\); when \(x=b\), \(t=1\). So, the new integral becomes: \[\int_{0}^{1} \frac{2t(b-a)dt}{\sqrt{((a + (b-a)t^2 -a)(b-(a + (b-a)t^2))}\]

Simplify and evaluate the integral

Simplify the expression above to: \[\int_{0}^{1} \frac{2t(b-a)dt}{\sqrt{(t^2(b-a))(b-(b-t^2(b-a)))}\] Now, cancel the \((b-a)\) terms and further simplify the integrand: \[\int_{0}^{1} \frac{2t}{\sqrt{t^2(1-t^2)}} dt\] Now, substitute \(u=t^2\), \(du = 2t dt\): \[\int_{0}^{1} \frac{du}{\sqrt{u(1-u)}}\] This integral evaluates to \(\pi\), which can be shown using a trig substitution or by comparing it to a known integral such as \(\int_{0}^{1} \frac{dx}{\sqrt{x(1-x)}}\). Thus, the first integral is proven. #2. Second substitution: \((b-x)(x-a)=t\)#

Substitute \((b-x)(x-a)=t\) in the second integral

For the second integral, we let \((b-x)(x-a)=t\). Then, we find the derivative to get \(-2x+(a+b)=-\frac{dt}{dx}\), and substitute these values into the integral: \[\int_{a}^{b} \frac{x\left(-\frac{1}{2x-(a+b)}\right) dx}{\sqrt{(x-a)(b-x)}}\] Change the limits of integration according to the new variable \(t\). When \(x=a\), \(t=0\); when \(x=b\), \(t=0\). The new integral becomes: \[\int_{0}^{0} \frac{\left(-\frac{1}{2x-(a+b)}\right) dx}{\sqrt{(x-a)(b-x)}}\]

Evaluate integral

Since the limits of integration are both 0, this integral is also 0. Now, we have to evaluate the same integral with the substitution \((b-x)(x-a)=t\) in the third equation to show that it is equal to \(\frac{1}{2}\pi (a+b)\). #3. Third substitution: \(x = a\cos^2t + b\sin^2t\)#

Substitute \(x = a\cos^2t + b\sin^2t\) in the second integral

For the third integral, use the substitution \(x = a\cos^2t + b\sin^2t\). Then, find \(dx = (b-a)(-2\sin t\cos t)dt\) and plug these values into the integral: \[\int \frac{(a\cos^2t + b\sin^2t)(-2\sin t\cos t (b-a)) dt}{\sqrt{((a\cos^2t + b\sin^2t -a)(b-(a\cos^2t + b\sin^2t))}\]

Simplify and evaluate the final integral

Simplify the expression above and recognize trigonometric identities (such as \(\sin^2t + \cos^2t = 1\) and \(2\sin t \cos t = \sin 2t\)): \[\int_{0}^{\frac{\pi}{2}}(a+b)\sin2t dt\] Evaluate the integral: \[\left.\frac{1}{2}(a+b)\cos 2t \right|_0^{\frac{\pi}{2}}\] Plug in the limits of integration and obtain: \[\frac{1}{2}\pi(a+b)\] Thus, the second integral is proven to be equal to \(\frac{1}{2}\pi (a+b)\).

Key concepts

Definite Integral

In Integral Calculus, the concept of a definite integral is fundamental. It represents the accumulation of quantities, and geometrically, it can be interpreted as the area under a curve bounded by a certain interval. The definite integral of a function f(x) from a to b is denoted as \[ \int_{a}^{b} f(x) dx \].

The calculation process involves finding an antiderivative (also known as an indefinite integral) of the function, which is a function F(x) such that F'(x) = f(x). The Fundamental Theorem of Calculus asserts that the definite integral from a to b can be found by evaluating the antiderivative at b and subtracting its value at a, which is mathematically expressed as \[F(b) - F(a)\].

This process can be seen in the provided textbook exercise, where the student is asked to evaluate an integral with given limits, and through substitution, transform it into a more familiar form that can be easily integrated.

Trigonometric Substitution

When an integral involves square roots or quadratic expressions, trigonometric substitution can be an effective technique to simplify the integral. It relies on using trigonometric identities to replace certain expressions with a trigonometric function of a new variable, which makes the integrand easier to handle.

An example of a trigonometric identity used in substitution is \(\sin^2(t) + \cos^2(t) = 1\), which stems from the Pythagorean Theorem. In the context of our exercise, the third substitution involves the expression \(x = a\cos^2(t) + b\sin^2(t)\), where the properties of sine and cosine functions help in simplifying the quadratic forms involving \(x\).

This not only streamlines the integration process but also can transform an integral that is very difficult (or even impossible) to solve in its original form into one that is readily solvable. The key to successfully applying trigonometric substitution lies in recognizing patterns in the integral that match trigonometric identities.

Integration by Substitution

The method of integration by substitution, also known as u-substitution, is a technique that transforms the integral of a composite function into an integral that is easier to evaluate. It is the reverse process of the Chain Rule in differentiation.

During this technique, a new variable \(u\) is introduced, which is a function of \(x\), and the differential \(dx\) is expressed in terms of \(du\). This change of variable simplifies the integral by making the integrand a function of \(u\) which is easier to integrate.

Steps for Integration by Substitution:

• Select a substitution \(u = g(x)\) that simplifies the integrand.
• Compute the differential \(du = g'(x)dx\).
• Change the limits of integration if dealing with a definite integral.
• Substitute \(u\) and \(du\) into the integral, replacing all \(x\) terms.
• Integrate with respect to \(u\).
• If necessary, convert back to the original variable \(x\) after integration.

Following these steps can enable students to work through challenging integrals by reducing them to simpler forms, as illustrated in the exercise where a simple substitution was first performed before applying trigonometric substitution.