Problem 42 - Calculus Single Variable

Question

Use trigonometric identities to compute the indefinite integrals. \(\text { Evaluate } \int \frac{1}{2^{x}} d x\)

Step-by-Step Solution

Answer

\(-\frac{1}{\ln 2} \, 2^{-x} + C\).

Step 1: Rewrite the Function

The given integral is \(\int \frac{1}{2^x} \, dx\). We start by rewriting \(\frac{1}{2^x}\) using exponent rules. Recall that \(2^x = e^{x \ln 2}\), thus \(\frac{1}{2^x} = 2^{-x} = e^{-x \ln 2}\). The integral then becomes \(\int e^{-x \ln 2} \, dx\).

Step 2: Integrate Using Substitution

To integrate \(\int e^{-x \ln 2} \, dx\), let \(u = -x \ln 2\), resulting in \(\frac{du}{dx} = -\ln 2\), or \(du = -\ln 2 \, dx\). Thus, \(dx = \frac{du}{-\ln 2}\), and the integral becomes \(\int e^u \cdot \frac{du}{-\ln 2}\).

Step 3: Simplify the Integral

The integral in terms of \(u\) is \(-\frac{1}{\ln 2} \int e^u \, du\). The integral of \(e^u\) is \(e^u\), hence the solution is \(-\frac{1}{\ln 2} \, e^u\).

Step 4: Substitute Back

Substitute back \(u = -x \ln 2\), giving \(-\frac{1}{\ln 2} \, e^{-x \ln 2}\). Since \(e^{-x \ln 2} = 2^{-x}\), the expression becomes \(-\frac{1}{\ln 2} \, 2^{-x}\).

Step 5: Add the Constant of Integration

Finally, add the constant of integration \(C\) to the indefinite integral, resulting in the solution: \(-\frac{1}{\ln 2} \, 2^{-x} + C\).

Key Concepts

Trigonometric Identities

Though not directly required in this particular problem of evaluating the integral of \(~\frac{1}{2^x} dx~\), trigonometric identities are incredibly useful when dealing with integrals that have trigonometric functions in them. Trigonometric identities allow us to transform complex trigonometric expressions into simpler ones that are more manageable to integrate. This often involves using identities like:

• Sum and difference identities: \(\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b\)
• Pythagorean identities: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Double angle identities: \(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta\)

By mastering these identities, you'll be able to tackle a wide range of integrals that may initially appear complicated but turn into simpler forms that are easier to solve.

Exponent Rules

Exponent rules are vital when dealing with integrals involving exponential functions. In the exercise, we saw that \(~\frac{1}{2^x}~\) can be rewritten using exponent rules as \(2^{-x}\), which further simplifies to \(e^{-x \ln 2}\). Exponent rules that are handy include:

• Product of powers: \(a^m \cdot a^n = a^{m+n}\)
• Quotient of powers: \(\frac{a^m}{a^n} = a^{m-n}\)
• Power of a power: \( (a^m)^n = a^{m \cdot n}\)

Understanding these rules allows us to rewrite expressions differently, simplifying integration problems. When faced with a base change like converting \(2^x \) to \((e^{x \ln 2})\), knowing how to apply the natural logarithm is crucial because it relates exponential and logarithmic functions closely.

Substitution Method

The substitution method is a technique used to make integration easier by changing the variable of integration. In our example, we replaced \(x\) with \(u = -x \ln 2\) to transform the integral into a simpler form. This method involves:

• Choosing a substitution: Identify a part of the integral that can be rewritten as a new variable \((u)\).
• Find \(du\): Compute the derivative of the substitution and multiply it by \(dx\).
• Replace \(dx\) with \(du\) : Substitute back into the integral.
• Integrate: Once in terms of \(u\), solve the simplified integral.
• Back-substitute: Replace \(u\) back with the original expression.

The substitution method is especially useful for integrals where a straightforward integration is difficult. It effectively reduces the complexity by temporarily modifying variables.

Constant of Integration

When evaluating indefinite integrals, you must always remember to add a constant of integration, \(C\). This is because indefinite integrals represent a family of functions, each differing by a constant. If you forget to include \(C\), you might miss out on representing all possible solutions. In our integral evaluation of \(~\int \frac{1}{2^x} \, dx~\), the final solution is given by

• \( -\frac{1}{\ln 2} \cdot 2^{-x} + C \)

This \(C\) accounts for any value that satisfies the antiderivative since taking the derivative of a constant results in zero. Never omit \(C\) when dealing with indefinite integrals, or your solution may be incomplete. The constant highlights that an indefinite integral does not have a unique solution unless initial conditions or limits are specified.