91Ó°ÊÓ

The substitution method is one of the most powerful techniques in integration. The idea is to simplify the integral by substituting part of the integrand with a new variable. This approach makes the function easier to integrate. In our exercise, we begin by expressing \(\tan u\) in terms of sine and cosine: \[ \tan u = \frac{\text{cos } u}{\text{sin } u} \] We then make a substitution, letting \(v = \text{sin } u\). The derivative of \(v\) with respect to \(u\) is \(dv = \text{cos } u \ du\) so we can replace \( \text{cos } u \ du\) in the integral. This substitution transforms the integral into a much simpler form: \[ \tan u \ du = \frac{1}{v} \ dv \] Finally, we integrate in terms of \(v\), and then substitute back the original variable to get the final answer.

Trigonometric integrals often require converting trigonometric functions into simpler forms for integration. In the given exercise, we did this by expressing \( \tan u \) as \( \frac{\text{cos } u}{\text{sin } u} \). This conversion allowed us to use the substitution method to simplify the integral. Trigonometric integrals can involve a wide variety of identities and substitutions. The goal is always to simplify the integrand to a form that is easier to integrate. It helps to be familiar with trig identities such as:

• \( \text{sin}^2 u + \text{cos}^2 u = 1 \)
• \( \tan u = \frac{\text{sin} u}{\text{cos} u } \)
• \( \text{sec }^2 u = 1 + \tan^2 u \)

These and other identities can be extremely useful in transforming and solving trigonometric integrals.

Logarithmic integration is frequently used when the integrand includes a fraction of the form \( \frac{1}{x} \). This is identified as a standard form since the antiderivative of \( \frac{1}{x} \) is \( \text{ln} |x| + C \). In our exercise, after performing the substitution and converting \( \tan u \ du = \frac{1}{v} \ dv \), we recognize the integral \[ \tan u \ du = \frac{1}{v} \ dv = \text{ln} |v| + C \] as a straightforward logarithmic form. Thus, we integrate to get \( \text{ln} |v| + C \). Finally, substitute \(v\) back into \( \text{sin } u\). The result is \[ \text{ln} |\text{sin } u| + C \] It's crucial to remember that the absolute value is used to ensure the argument of the logarithm is positive, accommodating all values of \( \text{sin} u \).