Question

For the following problems, use the substitutions \(\tan \left(\frac{x}{2}\right)=t, d x=\frac{2}{1+t^{2}} d t, \sin x=\frac{2 t}{1+t^{2}}\), and \(\cos x=\frac{1-t^{2}}{1+t^{2}}\). Given \(\tan \left(\frac{x}{2}\right)=t\), derive the formulas \(d x=\frac{2}{1+t^{2}} d t, \sin x=\frac{2 t}{1+t^{2}}\), and \(\cos x=\frac{1-t^{2}}{1+t^{2}} .\)

Short answer

The steps using substitution derive: \( dx = \frac{2}{1+t^2} dt \), \( \sin x = \frac{2t}{1+t^2} \), \( \cos x = \frac{1-t^2}{1+t^2} \).

Step-by-step solution

Understanding the Substitution

We start with the substitution \( t = \tan \left(\frac{x}{2}\right) \). This is a trigonometric substitution known as the T-substitution or Weierstrass substitution, useful in integrating trigonometric functions. Our goal is to derive the expressions for \( dx \), \( \sin x \), and \( \cos x \) based on this substitution.

Derive the Expression for dx

Given \( t = \tan \left(\frac{x}{2}\right) \), we find the derivative with respect to \( t \):\[ \frac{dt}{dx} = \frac{1}{2} \sec^2 \left(\frac{x}{2}\right) \]Using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \), we have \( \sec^2 \left(\frac{x}{2}\right) = 1 + t^2 \). Therefore:\[ dt = \frac{1}{2}(1 + t^2) dx \]Solving for \( dx \), we get the expression:\[ dx = \frac{2}{1+t^2} dt \]

Derive the Expression for sin x

Using the substitution, develop expressions for \( \sin x \) and \( \cos x \):1. Using \( t = \tan \left(\frac{x}{2}\right) \) and the half-angle identity \( \sin x = \frac{2 \tan \left(\frac{x}{2}\right)}{1 + \tan^2 \left(\frac{x}{2}\right)} \), substitute \( t \) to get:\[ \sin x = \frac{2t}{1+t^2} \]

Derive the Expression for cos x

Similarly, for cosine, use the identity:\[ \cos x = \frac{1 - \tan^2 \left(\frac{x}{2}\right)}{1 + \tan^2 \left(\frac{x}{2}\right)} \]Substitute \( t = \tan \left(\frac{x}{2}\right) \):\[ \cos x = \frac{1-t^2}{1+t^2} \]

Review the Derived Formulas

We have successfully derived the formulas:1. \( dx = \frac{2}{1+t^2} dt \)2. \( \sin x = \frac{2t}{1+t^2} \)3. \( \cos x = \frac{1-t^2}{1+t^2} \)These results show that the T-substitution transforms variables to simplify integration involving trigonometric functions.

Key concepts

Weierstrass Substitution

The Weierstrass substitution, also known as the T-substitution, is a powerful technique for integrating trigonometric functions. By replacing trigonometric expressions with rational functions, you can simplify complex integrals. The idea is simple: replace \( \tan\left(\frac{x}{2}\right) \) with \( t \). This substitution comes with new expressions for \( \sin x \) and \( \cos x \), which are expressed in terms of \( t \):

• \( \sin x = \frac{2t}{1+t^2} \)
• \( \cos x = \frac{1-t^2}{1+t^2} \)

Using the derivatives and identities of trigonometric functions, you derive the Weierstrass substitution to ease the process of integration. By making trigonometric terms more manageable, this technique is particularly useful when dealing with indefinite integrals where direct integration and substitution can be complicated.

Half-Angle Identities

Half-angle identities play a crucial role in deriving simpler expressions for trigonometric functions. These identities show how to express a trigonometric function of half an angle in terms of a square root involving non-trigonometric functions. For the T-substitution, the half-angle identities are used to express \( \sin x \) and \( \cos x \) in terms of \( \tan\left(\frac{x}{2}\right) \). For example:

• \( \sin x = \frac{2 \tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \)
• \( \cos x = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} \)

These transformations make it easier to perform operations such as differentiation and integration, as the trigonometric functions are then expressed in terms of just one variable, \( t \). These identities are fundamental not just in calculus, but also in simplifying formulas and calculations in various fields of science and engineering.

Derivative of Trigonometric Functions

Understanding the derivatives of trigonometric functions forms the basis for utilizing the Weierstrass substitution effectively. The T-substitution relies on the derivative of the tangent half-angle, \( t = \tan\left(\frac{x}{2}\right) \). Calculating this derivative is fundamental because it enables the transformation of the differential \( dx \).Through the relationship:\[ \frac{dt}{dx} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \]and knowing \( \sec^2\theta = 1 + \tan^2\theta \), we arrive at:\[ dt = \frac{1}{2}(1 + t^2) dx \]Solving for \( dx \), it becomes clear:\[ dx = \frac{2}{1+t^2} dt \]This manipulation shows how derivatives allow us to translate an expression involving \( x \) into one involving \( t \), making integration more approachable. Understanding and applying these derivatives are key parts of calculus, especially when working with non-polynomial functions.