Question

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution \(u=\tan (x / 2)\) or, equivalently, \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u .\) Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2} \quad \text { and } \quad \cos x=2 \cos ^{2} \frac{x}{2}-1$$.

Short answer

Question: Verify the given relations A, B, and C for the functions x = 2 * arctan(u), sin(x) = (2u)/(1 + u^2), and cos(x) = (1 - u^2)/(1 + u^2) respectively. Answer: The given relations A, B, and C have been verified successfully. For relation A, we found that the derivative of x with respect to u is dx/du = 2/(1+u^2), and hence dx = 2/(1+u^2) du. For relation B, we used the double-angle formula for sin(x) and the right-triangle diagram, and found that sin(x) = (2u)/(1 + u^2). For relation C, the use of the double-angle formula for cos(x) and the right-triangle diagram resulted in cos(x) = (1 - u^2)/(1 + u^2).

Step-by-step solution

Verify relation A by differentiating x = 2 * arctan(u)

To find the derivative: dx/du, we will differentiate x = 2 * arctan(u) with respect to u. Using the chain rule, we get: $$\frac{dx}{du} = \frac{d}{du} (2 \tan^{-1}(u)) = 2 * \frac{d}{du}(\tan^{-1}(u)) $$ Recall that the derivative of arctan(u) is: $$ \frac{d}{du}(\tan^{-1}(u)) = \frac{1}{1+u^2} $$ Hence, the derivative of x with respect to u is: $$\frac{dx}{du} = 2 * \frac{1}{1+u^2} $$ Now, multiplying by du, we get the desired relation A: $$dx = \frac{2}{1+u^2} du $$

Verify relation B using a right-triangle diagram and the double-angle formula

To verify relation B, let's consider a right triangle with angles x/2 and y. According to the definition of tangent function, we have: $$\tan(\frac{x}{2}) = \frac{u}{1} \Rightarrow u = \tan(\frac{x}{2}) $$ Now, we can use the identity sin(x) = 2 * sin(x/2) * cos(x/2), $$ \sin(x) = 2 * \sin(\frac{x}{2}) * \cos(\frac{x}{2}) $$ We know that sin(x/2) = u/(sqrt(1+u^2)) and cos(x/2) = 1/(sqrt(1+u^2)). Substituting these values in the equation, we will get: $$\sin(x) = 2 * \frac{u}{\sqrt{1+u^2}} * \frac{1}{\sqrt{1+u^2}} = \frac{2u}{1+u^2} $$ This verifies relation B.

Verify relation C using a right-triangle diagram and the double-angle formula

To verify relation C, let's use the identity cos(x) = 2 * cos^2(x/2) - 1, $$ \cos(x) = 2 * \cos^2(\frac{x}{2}) - 1 $$ We know that cos(x/2) = 1/(sqrt(1+u^2)). Substituting this value in the equation, we will get: $$\cos(x) = 2 * (\frac{1}{\sqrt{1+u^2}})^2 - 1 = 2 * \frac{1}{1+u^2} - 1 = \frac{1-u^2}{1+u^2} $$ This verifies relation C. Thus, all the given relations A, B, and C have been verified successfully.

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus that involves finding the function whose derivative is given. In many cases, the function that needs to be integrated involves trigonometric expressions. Trigonometric substitutions are one of the key techniques used to simplify such integrals. By using certain trigonometric identities or substitutions, we can convert a difficult integral into a simpler form that is easier to solve.

For example, when we see a square root of the form \( \sqrt{a^2 - x^2} \), a common technique is to substitute \( x = a \sin(\theta) \), which simplifies the square root expression. These substitutions can be quite powerful, as they utilize the periodic properties of trigonometric functions to transform the integrand into a more manageable form. In addition to simplifying the integrand, trigonometric substitutions often lead to integrals that can be solved using inverse trigonometric functions or other standard techniques of integration.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \arcsin(x) \), \( \arccos(x) \), and \( \arctan(x) \), are used to find the angle corresponding to a given trigonometric ratio. These functions are particularly useful when performing integrations, as they often appear as results of integrals involving trigonometric expressions. For instance, the integral of \( 1/(1+x^2) \) yields \( \arctan(x) \), which is exactly the substitution seen in the exercise.

It's critical to understand the properties and derivatives of these inverse functions to accurately perform integration using trigonometric substitutions. For example, the derivative of \( \arctan(u) \) is \( 1/(1+u^2) \) as demonstrated in the step-by-step solution provided, which is essential in verifying relation A in the exercise.

Double-Angle Formulas

Trigonometric double-angle formulas express trigonometric functions of an angle \( 2\theta \) in terms of \( \theta \). These are vital when decomposing more complex trigonometric expressions into simpler ones. The double-angle formulas for sine and cosine are \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \) and \( \cos(2\theta) = 2\cos^2(\theta) - 1 \) or \( \cos(2\theta) = 1 - 2\sin^2(\theta) \) respectively.

In the context of our exercise, relations B and C are verified using double-angle formulas by expressing \( \sin(x) \) and \( \cos(x) \) in terms of \( \sin(\frac{x}{2}) \) and \( \cos(\frac{x}{2}) \) respectively. This transformation enables us to relate the functions in terms of \( u \) by considering a right-triangle diagram and allows us to integrate expressions involving \( \sin(x) \) and \( \cos(x) \) after the appropriate substitution has been applied. Thus, understanding double-angle formulas is essential for tackling integrals that involve trigonometric substitution.