91Ó°ÊÓ

In the realm of integral calculus, trigonometric identities serve as crucial tools for simplifying expressions before integration. These identities allow us to rewrite trigonometric functions in alternative forms that may reveal a more straightforward path to finding the integral. For instance, in the original exercise, the trigonometric functions \(\cot x\) and \(\csc x\) are involved, which are not directly integrable without transformation.

Two essential identities we use here are:

• \(\cot x = \frac{\cos x}{\sin x}\)
• \(\csc x = \frac{1}{\sin x}\)

By expressing \(\cot^3 x\) and \(\csc^3 x\) using these identities, the integrand is rewritten in terms of sine and cosine, which are more manageable in the context of integration. This simplification is a key step because it translates the integral into a form where other advanced techniques like substitution can be utilized efficiently, as we will see in the following sections.

The substitution method is an integral calculus technique used to make integration more straightforward by transforming the variable of integration. Specifically, this method reduces complex integrals into simpler forms that are easier to solve, especially when dealing with composed functions.

In the current exercise, after transforming the problem using trigonometric identities, we choose \( u = \sin x \) as our substitution. This decision transforms the integral with respect to \( x \) into an expression with respect to \( u \). Given that \( du = \cos x \, dx \), we can reformulate the entire problem using \( u \) and \( du \), effectively converting the initial trigonometric integral into an algebraic one:

• The original integral becomes \( \int \frac{\cos^2 x \cdot \cos x \, dx}{u^6} \).
• After substitution, it simplifies to \( \int \frac{(1 - u^2) \cdot du}{u^6} \), leveraging the identity \( \cos^2 x = 1 - \sin^2 x = 1 - u^2 \).

Through such transformation, the complexity of trigonometric terms is skillfully managed, setting the stage for uncomplicated integration.

Integral calculus is the branch of mathematics dealing with the determination of integrals and their applications. In the context of this exercise, we utilize the power of integral calculus to evaluate the integral after substitution and simplification of trigonometric terms.

Following substitution, our task boils down to integrating the expression \( \int (u^{-6} - u^{-4}) \, du \). This translates into finding the antiderivative of each term separately:

• For \( \int u^{-6} \, du \), the antiderivative is \( \frac{u^{-5}}{-5} = -\frac{1}{5u^5} \).
• In the case of \( \int u^{-4} \, du \), the resulting antiderivative is \( \frac{u^{-3}}{-3} = -\frac{1}{3u^3} \).

After evaluating these integrals, we combine them and finally revert the substitution back to our original variable, \( \sin x \). Therefore, the solution to the initial integral is expressed as \[ -\frac{1}{5\sin^5 x} + \frac{1}{3\sin^3 x} + C \],where \( C \) represents the constant of integration. This process highlights the beauty and efficiency of integral calculus in solving complex mathematical problems.