Question

Different substitutions a. Show that \(\int \frac{d x}{\sqrt{x-x^{2}}}=\sin ^{-1}(2 x-1)+C\) using cither \(u=2 x-1\) or \(u=x-\frac{1}{2}\) b. Show that \(\int \frac{d x}{\sqrt{x-x^{2}}}=2 \sin ^{-1} \sqrt{x}+C\) using \(u=\sqrt{x}\) c. Prove the identity \(2 \sin ^{-1} \sqrt{x}-\sin ^{-1}(2 x-1)=\frac{\pi}{2}\)

Short answer

In this exercise, we used two different substitution methods to evaluate the integral \(\int\frac{dx}{\sqrt{x-x^2}}\). In part (a), we used the substitution \(u=2x-1\) and found the integral to be \(\sin^{-1}(2x-1) + C\). In part (b), we used the substitution \(u=\sqrt{x}\) and found the integral to be \(2\sin^{-1}\sqrt{x} + C\). In part (c), we used the sine addition formula and the Pythagorean identity to prove the given identity: \(2\sin^{-1}\sqrt{x} - \sin^{-1}(2x-1) =\frac{\pi}{2}\). Overall, this exercise showcased the use of substitution methods and trigonometric properties in solving integrals and proving identities.

Step-by-step solution

Part (a) - Using substitution \(u=2x-1\)

We are given the substitution \(u=2x-1\). First, find the derivative of this substitution with respect to \(x\): $$\frac{du}{dx} = 2.$$ Now, solve for \(dx\): $$dx = \frac{1}{2} du.$$ Also, solve for \(x\) in terms of \(u\): $$x = \frac{1+u}{2}.$$ Now, replace \(x\) and \(dx\) in the integral: $$\int \frac{d x}{\sqrt{x-x^{2}}} = \int \frac{\frac{1}{2}du}{\sqrt{\frac{1+u}{2}-\left(\frac{1+u}{2}\right)^2}}.$$ Simplifying the integral, we get: $$\int \frac{du}{\sqrt{u-u^2}}.$$ Now, take the integral: $$\int \frac{du}{\sqrt{u-u^2}} = \sin^{-1}(u) + C.$$ Recall that \(u=2x-1\). Substitute this back in to get the answer for part (a): $$\sin^{-1}(2x-1) + C.$$

Part (b) - Using substitution \(u=\sqrt{x}\)

We are given the substitution \(u=\sqrt{x}\). First, find the derivative of this substitution with respect to \(x\): $$\frac{du}{dx} = \frac{1}{2\sqrt{x}}.$$ Now, solve for \(dx\): $$dx = 2\sqrt{x} du.$$ Also, solve for \(x\) in terms of \(u\): $$x = u^2.$$ Now, replace \(x\) and \(dx\) in the integral: $$\int \frac{dx}{\sqrt{x-x^2}} = \int \frac{2\sqrt{u^2} du}{\sqrt{u^2-{u^4}}}.$$ Simplifying the integral, we get: $$\int \frac{2 du}{\sqrt{1-u^2}}.$$ Now, take the integral: $$\int \frac{2 du}{\sqrt{1-u^2}} = 2\sin^{-1}(u) + C.$$ Recall that \(u=\sqrt{x}\). Substitute this back in to get the answer for part (b): $$2\sin^{-1}\sqrt{x} + C.$$

Part (c) - Proving the identity

We need to prove the identity: $$2\sin^{-1}\sqrt{x} - \sin^{-1}(2x-1) =\frac{\pi}{2}.$$ To prove this identity, we can make use of the sine addition formula: $$\sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta.$$ Set \(\alpha = 2\sin^{-1}\sqrt{x}\) and \(\beta = \sin^{-1}(2x-1)\). Now, find \(\sin(\alpha -\beta)\) using the sine addition formula: $$\sin(\alpha - \beta) = \sin\left(2\sin^{-1}\sqrt{x}\right)\cos\left(\sin^{-1}(2x-1)\right) - \cos\left(2\sin^{-1}\sqrt{x}\right)\sin\left(\sin^{-1}(2x-1)\right).$$ We want to show that \(\sin(\alpha - \beta) = 1\), which means \(\alpha - \beta = \frac{\pi}{2}\). To find \(\sin\left(2\sin^{-1}\sqrt{x}\right)\), use the double angle formula for sine: $$\sin\left(2\sin^{-1}\sqrt{x}\right) = 2\sin\left(\sin^{-1}\sqrt{x}\right)\cos\left(\sin^{-1}\sqrt{x}\right) = 2\sqrt{x}\sqrt{1-x}.$$ We can use the Pythagorean identity \(\cos^2(x)=1-\sin^2(x)\) to find the cosine for the inverse sine: $$\cos\left(\sin^{-1}(2x-1)\right)=\sqrt{1-\sin^2\left(\sin^{-1}(2x-1)\right)}=\sqrt{1-(2x-1)^2}.$$ Similarly, we can find the cosine for the inverse sine using the double angle formula for cosine: $$\cos\left(2\sin^{-1}\sqrt{x}\right)=\cos^2\left(\sin^{-1}\sqrt{x}\right)-\sin^2\left(\sin^{-1}\sqrt{x}\right)=1-2x.$$ Now, plug in these expressions into the expression for \(\sin(\alpha-\beta)\): $$\sin(\alpha-\beta) = 2\sqrt{x}\sqrt{1-x}\sqrt{1-(2x-1)^2} - (1-2x)(2x-1).$$ Upon simplification, we get: $$\sin(\alpha-\beta) = 1.$$ Hence, \(\alpha - \beta = \frac{\pi}{2}\), which concludes the proof of the identity: $$2\sin^{-1}\sqrt{x} - \sin^{-1}(2x-1) =\frac{\pi}{2}.$$

Key concepts

Indefinite Integrals

An indefinite integral, also known as an antiderivative, represents a family of functions that generate a given function as their derivative. It plays a fundamental role in calculus, particularly in the process of integration. When we write an indefinite integral, it comes with a constant of integration, often denoted as 'C', because differentiating the original function does not reveal any constant term.

For example, when solving the integral \[ \int \frac{dx}{\sqrt{x - x^2}} \] we are looking for all functions whose derivative gives us \(\frac{1}{\sqrt{x - x^2}}\). In the substitutions provided in the exercise, the goal is to manipulate the integral into a form that we recognize and can easily integrate, either by a simple power rule or a standard integral such as \(\sin^{-1}(x)\).

Exercise Improvement Advice:

• To grasp the concept of indefinite integrals, it is essential to focus on the method of substitution which simplifies the integration process.
• Practicing variation in substitution methods can help understand how different substitutions can lead to the same result, reinforcing the importance of recognizing integral forms.

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial for solving integrals involving square roots of expressions similar to \(1 - x^2\) or \(x^2 - 1\). They are the inverses of the trigonometric functions such as sine, cosine, and tangent, and include \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\). These functions return the angle whose trigonometric function yields the value x.

In the exercise, the use of \(\sin^{-1}(x)\) appears because the integral of \(\frac{1}{\sqrt{1 - u^2}}\) is the arcsine of u. It is vital to familiarize oneself with the derivatives and integrals of inverse trigonometric functions, as they often are the result of applying integration techniques like substitution.

Exercise Improvement Advice:

• Understanding the relationship between trigonometric functions and their inverses is key when simplifying integral expressions.
• Recognizing the derivative of an inverse trigonometric function can aid in identifying when to use it for integration.

Trigonometric Identities

Trigonometric identities are equalities involving trigonometric functions that hold true for all admissible values of the variables they contain. They can simplify complex expressions and solve trigonometric equations. Some of the most commonly used identities include Pythagorean identities, angle sum and difference identities, and double angle formulas.

In the given exercise, the trigonometric identity used is the sine addition formula, which simplifies the expression and allows us to verify the identity by comparing sides. In calculus, knowing these identities by heart is instrumental in manipulating integrals to favorable forms for calculation. For example, knowing that \(\cos^2(x) = 1 - \sin^2(x)\) allows us to find the cosine of an inverse sine without directly computing an angle.

Exercise Improvement Advice:

• Regular practice of trigonometric identities is crucial for their application in calculus exercises involving integration and differentiation.
• Understanding how the identities are derived can facilitate their recall and utility in various mathematical scenarios.