Problem 65 - Calculus Early Transcendentals

Question

Use a change of variables to evaluate the following integrals. $$\int \frac{\csc ^{2} x}{\cot ^{3} x} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $\int \frac{\csc ^{2} x}{\cot ^{3} x} d x$. Answer: $-\frac{1}{2\cos^2 x} - 2\ln |\cos x| + \frac{1}{2}\cos^2 x + C$

Step 1: Rewrite the expressions in terms of sine and cosine

Since \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\), rewrite the integrand as: $$\int \frac{\csc ^{2} x}{\cot ^{3} x} dx =\int \frac{(\frac{1}{\sin{x}})^2}{(\frac{\cos{x}}{\sin{x}})^3} dx =\int \frac{\sin^3 x}{\cos^3 x} dx$$

Step 2: Apply u-substitution

Let \(u = \cos x\) so \(-du = \sin x \, dx\). Therefore, the integral becomes: $$ \int \frac{\sin^3 x}{\cos^3 x} dx = -\int \frac{(1-u^2)^2}{u^3} (-du) = \int \frac{(1-u^2)^2}{u^3} du $$

Step 3: Expand the numerator

Expand the numerator in the integrand: $$\int \frac{(1-u^2)^2}{u^3} du=\int \frac{1-2u^2+u^4}{u^3} du$$

Step 4: Split the integral

Split the integral into three separate integrals: $$\int \frac{1-2u^2+u^4}{u^3} du = \int\left(\frac{1}{u^3}-\frac{2u^2}{u^3}+\frac{u^4}{u^3}\right)du = \int\left(\frac{1}{u^3}-\frac{2}{u}+\frac{u}{u}\right)du=\int \frac{1}{u^3}du - \int 2\frac{1}{u}du + \int u du$$

Step 5: Integrate and substitute back

Integrate each integral and then substitute back for \(x\) using \(u = \cos x\): $$\int \frac{1}{u^3} du =- \frac{1}{2u^2} +C_1$$ $$\int 2\frac{1}{u} du =- 2 \ln |u| + C_2$$ $$\int u du =\frac{1}{2}u^2 + C_3$$ Substitute back \(u = \cos x\), so the final answer is: $$-\frac{1}{2\cos^2 x} - 2\ln |\cos x| + \frac{1}{2}\cos^2 x + C$$ where \(C = C_1 + C_2 + C_3\) is the constant of integration.

Key Concepts

Change of Variables

When you face complex integrals, changing variables can simplify your work. This method allows you to replace a variable with a new one that often eases the calculations.
Instead of directly integrating the original function, we look for a new variable that can transform the integral into a simpler form.
In our example, we first expressed everything in terms of sine and cosine, making it easier to handle.
After simplifying the given integral, we made it easier using a new variable, u, instead of the original functions. This variable change helped break down the integral into simpler parts.

• Identify expressions within the integral that can be simplified more easily.
• Determine an appropriate new variable that will simplify these expressions.
• Transform the entire integral, including the differential, by substituting the new variable.

This technique is particularly helpful when the function involves complex algebraic or trigonometric expressions.

Trigonometric Functions

Trigonometric functions often appear in integrals, but they can initially look intimidating. Knowing their identities can significantly help transform and simplify integrals.
In our problem, the trigonometric identities such as \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{\cos x}{\sin x}\) were used to rewrite the original integral.
This conversion from reciprocal trigonometric functions into standard sine and cosine transforms the integral into a more manageable form.

• Reciprocal identities help convert expressions into sine and cosine which are easier to manipulate.
• Product-to-Sum formulas and Pythagorean identities can reduce complex expressions or simplify integrands.

Using a strong understanding of trigonometric identities makes handling these functions much more straightforward, allowing you to focus on integration rather than being stuck on transformations.

u-substitution

U-substitution is a powerful technique that allows you to transform complex integrands into simpler ones.
It involves setting part of the integral as a new variable, \(u\), which simplifies the integration process significantly.
In the given problem, choosing \(u = \cos x\) allowed the integral to be broken into more manageable parts. This substitution also requires changing \(dx\) into terms of \(du\), ensuring the entire integral is expressed in terms of \(u\).

• Identify the part of the integrand that resembles a known derivative.
• Express \(dx\) in terms of \(du\), correctly incorporating any constants.
• After integration, substitute back for the original variable to give the solution in terms of the original variables.

This method is intuitive once you determine the right expression to substitute and is immensely helpful in solving otherwise complex integrals.