Problem 18 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int \cot ^{2} x \csc ^{2} x d x$$.

Step-by-Step Solution

Answer

The short answer to the integral \(\int \cot ^{2} x \csc ^{2} x d x\) is \(-\frac{1}{3}(\cot(x))^3 + C\).

Step 1: Recall the relationship between cotangent and cosecant functions and their derivatives

Remember the following derivatives: \(\\ \frac{d}{d x}\cot(x) = -\csc^2(x) \\ \\ \frac{d}{d x}\csc(x) = -\csc(x)\cot(x) \)

Step 2: Perform a substitution

Let \(u = \cot(x)\). Then, the derivative of \(u\) with respect to \(x\) is: \(\frac{du}{dx} = -\csc^2(x)\) Now, rewrite the given integral in terms of \(u\): \(\int \cot^2(x)\csc^2(x) dx = \int u^2(-du)\)

Step 3: Integrate the expression in terms of u

Now, we can integrate with respect to \(u\): \(\int u^2(-du) = -\frac{1}{3} u^3 + C\)

Step 4: Substitute back in terms of x

Finally, substitute the original function of \(x\) back into the equation: \(-\frac{1}{3}(\cot(x))^3 + C\)

Step 5: Write the final answer

The integral of the given function is: \(\int \cot^2(x)\csc^2(x) dx = -\frac{1}{3}(\cot(x))^3 + C\)

Key Concepts

Differential Calculus

Differential calculus is all about calculating derivatives, which tell us how a function changes as its input changes. Think of this as figuring out the slope of a curve at any point. For trigonometric functions like sine, cosine, tangent, and their counterparts (like cotangent and cosecant), each has a specific derivative. These derivatives help us understand how the functions behave and change.

• The derivative of \(\cot(x)\) is \(-\csc^2(x)\)\, useful when dealing with integrals involving \(\cot(x)\).
• The derivative of \(\csc(x)\) is \(-\csc(x)\cot(x)\), reflecting how closely related trigonometric functions are.

Understanding these derivatives is crucial for performing substitutions in integration, essentially reversing the differentiation process. In the exercise's solution, knowing that \(\frac{d}{dx}\cot(x) = -\csc^2(x)\) allowed us to express the integral in terms of another variable \(u\), which made the integration process easier.

Trigonometric Integrals

Trigonometric integrals are integrals that contain trigonometric functions such as sine, cosine, tangent, etc. Solving these integrals often requires the use of substitution or algebraic identities that simplify the process.

When integrating functions with trigonometric terms, like in our problem with \(\cot^2(x)\csc^2(x)\), knowing trigonometric relationships and identities can help. For example, we used:

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

Such identities enable us to express the trigonometric integral in a more manageable form. When trigonometric expressions become complex, substitution helps turn the problem into something more straightforward, like a polynomial, which is easier to integrate.

Antiderivatives

Antiderivatives, or indefinite integrals, are the reverse process of finding derivatives. When you integrate a function, you find another function whose derivative is the original function you started with. In our example, finding the antiderivative of \(\cot^2(x)\csc^2(x)\) involved substituting \(\cot(x)\) with \(u\). Once we did that, the integral became a simple polynomial in terms of \(u\): \(u^2\).

The integration of \(u^2\) is straightforward, resulting in:

• \(-\frac{1}{3}u^3 + C\)

This new expression represents the antiderivative of our original integral problem, expressed in terms of \(u\). Substituting back \(\cot(x)\) for \(u\) gives us \(-\frac{1}{3}(\cot(x))^3 + C\). Remember: the \(+C\) term is important because it represents the constant of integration, accounting for any constant that could have been differentiated away.