91Ó°ÊÓ

Trigonometric substitution is a method used in calculus to simplify integrals that involve trigonometric functions. When faced with a complex integral such as \( \int \cos \theta \cos^{5}(\sin \theta) \, d\theta \), identifying its type is key. Here, the functions \( \cos \theta \) and \( \sin \theta \) are wrapped in a composition. This suggests using a substitution.

In this case, the substitution \( u = \sin \theta \) is employed. This choice is strategic: it leverages the relationship between \( \sin \theta \) and its derivative \( \cos \theta \), as \( \frac{du}{d\theta} = \cos \theta \). Hence, the differential \( du \) becomes exactly \( \cos \theta \, d\theta \). By substituting, the original integral reduces to a simpler form \( \int u^5 \, du \). This process of substitution transforms a complicated problem into a much easier one.

Power rule integration is an essential technique in calculus for integrating functions of the form \( u^n \). When the function inside an integral matches this pattern, the power rule applies, making integration straightforward.

In this exercise, after substituting \( u = \sin \theta \), the integral transformed to \( \int u^5 \, du \). Using the power rule, we integrate \( u^n \) by calculating \( \frac{u^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

• For \( n = 5 \), this becomes \( \frac{u^{6}}{6} + C \).

This simplification speeds up solving integrals and reduces the risk of errors, focusing solely on the power of the variable.

While the exercise question involves an indefinite integral, understanding the concept of definite integrals is vital too. A definite integral computes the accumulated value of a function over a specific interval \([a, b]\).

In contrast to indefinite integrals, which include an arbitrary constant \( C \), definite integrals produce actual numerical values, often representing areas under curves.

• If the problem specified limits of integration, like \( \int_{a}^{b} \), you'd evaluate the antiderivative at these bounds.
• This process involves plugging in the upper and lower limits after finding the antiderivative to get \( F(b) - F(a) \).

Therefore, while indefinite integrals focus on antiderivatives, definite integrals are crucial for calculations involving ranges and total values over intervals.