Problem 30 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int \frac{\sin ^{3} x}{\cos x} d x$$.

Step-by-Step Solution

Answer

The integral of \(\frac{\sin ^{3}{x}}{\cos{x}}\) with respect to x can be calculated as follows: 1. Rewrite the integrand using the trigonometric identity \(\sin ^{2}{x} = 1 - \cos ^{2}{x}\) and split the integral into two parts. 2. Simplify the resulting integrals to get \(\int \tan{x} dx\) and \(\int \sin{x} \cos{x} dx\). 3. Integrate the first integral using substitution: \(-\ln |\cos{x}| + C_{1}\). 4. Integrate the second integral using integration by parts: \(\frac{\sin ^{2}{x}}{2} - \frac{1}{4} \cos{2x} + C_{2}\). 5. Combine the results to obtain the final answer: \(-\ln |\cos{x}| + \frac{\sin ^{2}{x}}{2} - \frac{1}{4} \cos{2x} + C\).

Step 1: Simplify the expression using trigonometric identities

We know that $$\sin ^{2}{x} = 1 - \cos ^{2}{x}$$. Using this identity, we can rewrite the integrand as: $$\frac{\sin{x} (\sin ^{2}{x})}{\cos{x}} = \frac{\sin{x} (1 - \cos ^{2}{x})}{\cos{x}}$$ Now our integrand becomes $$\int \frac{\sin{x} (1 - \cos ^{2}{x})}{\cos{x}} dx$$.

Step 2: Split the integral into two parts

To further simplify the expression, we can split the integral into two parts: $$\int \frac{\sin{x} (1 - \cos ^{2}{x})}{\cos{x}} dx = \int \frac{\sin{x}}{\cos{x}}dx - \int \frac{\sin{x} \cos ^{2}{x}}{\cos{x}} dx$$

Step 3: Simplify the resulting integrals

We'll now simplify the resulting integrals: $$\int \frac{\sin{x}}{\cos{x}}dx = \int \tan{x} dx$$ $$\int \frac{\sin{x} \cos ^{2}{x}}{\cos{x}} dx = \int \sin{x} \cos{x} dx$$

Step 4: Integrate the first part using substitution

For the first integral, we'll perform a substitution: Let $$u = \cos{x}$$, so $$\frac{du}{dx} = -\sin{x}$$. We can write the integral as: $$\int \tan{x} dx = -\int \frac{1}{u} du$$ Now we can integrate with respect to u: $$-\int \frac{1}{u} du = -\ln |u| + C_{1} = -\ln |\cos{x}| + C_{1}$$

Step 5: Integrate the second part using integration by parts

For the second integral, we'll use integration by parts: Let $$v = \sin{x}$$ and $$dw = \cos{x} dx$$, so $$dv = \cos{x} dx$$ and $$w = \sin{x}$$. Now we have: $$\int \sin{x} \cos{x} dx = \int v dw = vw - \int w dv$$ Substitute the values of v and w: $$= (\sin{x})(\sin{x}) - \int (\sin{x})(\cos{x}) dx$$ $$= \frac{\sin ^{2}{x}}{2} - \frac{1}{2} \int \sin{2x} dx$$ Now, we can use substitution again for the remaining integral: Let $$z = 2x$$, so $$\frac{dz}{dx} = 2$$, and $$\frac{dz}{2} = dx$$ So our integral becomes: $$\frac{1}{2} \int \sin{z} \frac{dz}{2} = \frac{1}{4} \int \sin{z} dz$$ Integrating with respect to z, we get: $$-\frac{1}{4} \cos{z} + C_{2}= -\frac{1}{4} \cos{2x} + C_{2}$$

Step 6: Add the two results to find the final answer

Combine the results from Step 4 and Step 5 to obtain the final answer: $$\int \frac{\sin ^{3}{x}}{\cos{x}}dx = -\ln |\cos{x}| + \frac{\sin ^{2}{x}}{2} - \frac{1}{4} \cos{2x} + C$$

Key Concepts

Integration by Substitution

When dealing with complex integrals, integration by substitution is a powerful technique that simplifies the process. This method involves substituting a part of the integrand with a new variable to transform the integral into an easier form. In our exercise, we apply substitution in two key steps.

First, when tackling the integral \( \int \tan{x} dx \), we substitute \( u = \cos{x} \) leading to \( \frac{du}{dx} = -\sin{x} \). This transforms the integral into \( -\int \frac{1}{u} du \), which is much simpler to integrate, resulting in \( -\ln |u| \) or \( -\ln |\cos{x}| \) when reverting to the original variable.

Using substitution helps to break down the complex trigonometric functions into more manageable expressions. The key is to identify a substitution that simplifies the integrand and makes it easier to evaluate. Remember to always change the limits of integration if it's a definite integral, or revert back to the original variable if it's an indefinite one, as shown in the exercise.

Integration by Parts

In calculus, integration by parts is essential when dealing with products of functions. The method stems from the product rule for differentiation and is useful when one term in the integrand can easily become a derivative. The formula is based on: \[ \int u \, dv = uv - \int v \, du \]

In our solution, integration by parts is applied to the integral \( \int \sin{x} \cos{x} dx \). We picked \( v = \sin{x} \) which after differentiation gives \( dv = \cos{x} \, dx \), and \( dw = \cos{x} \, dx \) giving \( w = \sin{x} \) after integration. Substituting these into the formula, we obtain: \( (\sin{x})(\sin{x}) - \int (\sin{x})(\cos{x}) \, dx \).

This simplifies further by recognizing the pattern, leading us to break down the remaining expression into simpler parts that can be solved using substitution. Apply the technique when facing difficult integrals involving products. It may take several steps, but the method will turn complicated integrals into solvable ones.

Trigonometric Identities

Trigonometric identities are fundamental relationships that express trigonometric functions in terms of one another. These identities play a crucial role in simplifying integrals involving trigonometric functions.

In this exercise, we utilized the identity \( \sin^2{x} = 1 - \cos^2{x} \) to rewrite \( \sin^3{x} \) within the integral \( \int \frac{\sin^3{x}}{\cos{x}} dx \). By expressing \( \sin^2{x} \) in terms of \( \cos{x} \), the original trigonometric expression is simplified, allowing for easier integration using substitution and integration by parts.

This demonstrates that knowing and applying the right trigonometric identity can transform complex expressions into simpler forms. Being familiar with identities such as the Pythagorean identities, double angle formulas, and half-angle formulas is indispensable for efficiently solving integrals involving trigonometric functions. These identities are your toolbox for tackling a broad range of calculus problems.