91Ó°ÊÓ

Trigonometric substitution is a valuable technique in calculus, particularly useful for solving integrals involving trigonometric functions. In the given problem, we apply the substitution \( u = \tan(x/2) \). This substitution is useful because it transforms the integral into a more manageable form.

With this approach, the trigonometric functions \( \sin x \) and \( \cos x \) are expressed in terms of \( u \):

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

From these identities, we derive \( \sec x = \frac{1+u^2}{1-u^2} \). This clever substitution simplifies the differentiation and the integration process, allowing for easier manipulation of the expression.

By substituting \( u \) into our integral, we also need to adjust \( dx \). Differentiating \( x = 2 \arctan(u) \) yields \( dx = \frac{2}{1+u^2} \, du \). Using this differential expression, the integral transforms into a simpler form, making it solvable by other techniques like partial fraction decomposition.

Partial fraction decomposition is a technique used to express a fraction as a sum of simpler fractions. In our problem, after making a trigonometric substitution, the integral becomes \( 2 \int \frac{1}{1-u^2} \, du \). This form is suitable for partial fraction decomposition.

The expression \( \frac{1}{1-u^2} \) can be decomposed as:

• \( \frac{1}{1-u^2} = \frac{1}{2} \left( \frac{1}{1+u} + \frac{1}{1-u} \right) \)

This allows us to integrate each fraction separately. By integrating, we get \( \ln|1+u| - \ln|1-u| + C \), which simplifies to \( \ln|\frac{1+u}{1-u}| + C \) after combining the logarithms.

This process highlights how partial fraction decomposition simplifies the integrals of rational expressions resulting from trigonometric substitutions, making complex integrals easier to solve.

Trigonometric identities are essential tools in calculus and trigonometry, often used to simplify or transform expressions. Part (b) of the problem utilizes the identity for the tangent of a sum: \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).

Here, we set \( A = \frac{\pi}{4} \) and \( B = \frac{x}{2} \), so:

• \( \tan(\frac{\pi}{4}) = 1 \)
• \( \tan(\frac{\pi}{4} + \frac{x}{2}) = \frac{1 + \tan(x/2)}{1 - \tan(x/2)} \)

This identity is used to re-express the integral in a form that matches the problem statement from part (a).

Using this identity, we find that the expression \( \ln \left| \frac{1 + \tan(x/2)}{1 - \tan(x/2)} \right| \) can also be rewritten as \( \ln \left| \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| \). This confirms the consistency and correctness of both expressions, tying the problem together with a neat application of trigonometric identities.