Problem 48 - Calculus Single Variable

Question

Evaluate the given integral by converting the integrand to an expression in sines and cosines. $$ \int \tan ^{3}(x / 2) d x $$

Step-by-Step Solution

Answer

The evaluated integral is \(-\sec^2(x/2) + 2 \ln|\cos(x/2)| + C\).

Step 1: Rewrite Tangent in Terms of Sine and Cosine

First, we express \( \tan(x/2) \) in terms of sine and cosine using the identity \( \tan(u) = \frac{\sin(u)}{\cos(u)} \). Therefore, \( \tan(x/2) = \frac{\sin(x/2)}{\cos(x/2)} \). Consequently, \( \tan^3(x/2) = \left(\frac{\sin(x/2)}{\cos(x/2)}\right)^3 = \frac{\sin^3(x/2)}{\cos^3(x/2)} \).

Step 2: Substitution

To simplify the integration, we use a substitution method by letting \( u = x/2 \), which implies \( dx = 2\, du \). The integral becomes \( 2 \int \frac{\sin^3(u)}{\cos^3(u)} du \).

Step 3: Express \( \sin^3(u) \) Using \( \sin(u) \) and \( \cos(u) \)

We know that \( \sin^3(u) = \sin(u) \cdot \sin^2(u) \). Using \( \sin^2(u) = 1 - \cos^2(u) \), express \( \sin^3(u) = \sin(u)(1 - \cos^2(u)) = \sin(u) - \sin(u)\cos^2(u) \).

Step 4: Decomposition of the Integral

We can further decompose the integral: \( 2 \int \frac{\sin(u)}{\cos^3(u)} du - 2 \int \frac{\sin(u)\cos^2(u)}{\cos^3(u)} du \). Simplifying the second term gives \(-2 \int \frac{\sin(u)}{\cos(u)} du \).

Step 5: Simplify and Integrate Each Part Separately

First, solve \( 2 \int \frac{\sin(u)}{\cos^3(u)} du = 2 \int \frac{\sin(u)}{\cos(u) \cdot \cos^2(u)} du \). Use substitution here with \( v = \cos(u) \), \( dv = -\sin(u) du \). This gives \(-2 \int v^{-3} dv = -\frac{2}{2} v^{-2} = -\frac{1}{\cos^2(u)} \).For the second part, \( -2 \int \frac{\sin(u)}{\cos(u)} du = -2 \int d(-\ln|\cos(u)|) = 2 \ln|\cos(u)| \).

Step 6: Combine Results and Back Substitute

Combine the results to get \(-\sec^2(u) + 2 \ln|\cos(u)| \) and substitute back \( u = x/2 \). Therefore, the evaluated integral is \(-\sec^2(x/2) + 2 \ln|\cos(x/2)| + C \), where \( C \) is a constant of integration.

Key Concepts

Trigonometric Integrals

Trigonometric integrals are special types of integrals where the integrand involves trigonometric functions such as sine, cosine, and tangent. When dealing with such integrals, certain approaches can simplify the expression and make integration easier.

• One common technique is to convert all trigonometric functions involved into sines and cosines. This conversion helps unify the trigonometric functions allowing for simpler manipulation.
• In cases like \( \int \tan^3(x/2) \, dx \), expressing tangent in terms of sine and cosine can lead to rewriting the integrand as \( \frac{\sin^3(x/2)}{\cos^3(x/2)} \), making it more amenable to further integration techniques.

By understanding the characteristics of trigonometric integrals and utilizing identities, solving these can become less daunting.

U-Substitution

U-substitution is a common technique used in integration which is similar to reversing the chain rule for differentiation. It involves changing the variable of integration to simplify the integrand.

• Here's how it works: choose a substitution \( u = g(x) \) such that the derivative \( g'(x) \) appears in the integral, allowing you to rewrite the differential \( dx \) in terms of \( du \).
• For the integral \( \int \tan^3(x/2) \, dx \), by setting \( u = x/2 \), we effectively reduce the complexity of the integral to \( 2 \int \frac{\sin^3(u)}{\cos^3(u)} \, du \).

This change simplifies calculations and is especially helpful when dealing with trigonometric integrals that involve complex expressions.

Trigonometric Identities

Trigonometric identities are mathematical tools that express relationships between different trigonometric functions, allowing us to transform expressions for ease of integration.

• Key identities include \( \sin^2(u) + \cos^2(u) = 1 \), which often comes in handy when expressions containing \( \sin^2(u) \) or \( \cos^2(u) \) need to be simplified.
• In the exercise, use \( \sin^2(u) = 1 - \cos^2(u) \) to rewrite \( \sin^3(u) = \sin(u)(1 - \cos^2(u)) = \sin(u) - \sin(u) \cos^2(u) \).

Understanding these fundamental identities is crucial for addressing complicated integrals involving trigonometric functions.

Integration by Parts

Integration by parts is a technique derived from the product rule for differentiation. It's particularly useful when the integrand is a product of two functions that don't simplify easily.

• The formula is \( \int u \, dv = uv - \int v \, du \). Choosing the right parts is key: \( u \) should be something that simplifies when differentiated, while \( dv \) should be easily integrable.
• In the given exercise, we realize that decomposing the integral into parts makes use of this method even without explicitly applying the integration by parts formula. The separation into simpler components allows for substitution that mimics using integration by parts.

Mastering this technique enables broader handling of complex integrals where other methods may be insufficient.