Evaluating The Integral Of Sin^3(x) / Cos(x) From 0 To Π/6

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Introduction

In this article, we will delve into the evaluation of a definite integral involving trigonometric functions. Specifically, we aim to find the value of the integral $\int_0^{\pi / 6} \frac{\sin ^3(x)}{\cos (x)} d x$. This problem combines trigonometric identities and substitution techniques, making it a compelling exercise in calculus. Understanding how to tackle such integrals is crucial for students and professionals in mathematics, physics, and engineering. Let's embark on this mathematical journey and break down the solution step by step.

Understanding the Integral

The given integral is a definite integral, which means we are looking for the area under the curve of the function $f(x) = \frac{\sin ^3(x)}{\cos (x)}$ between the limits $x = 0$ and $x = \frac{\pi}{6}$. The integrand, $\frac{\sin ^3(x)}{\cos (x)}$, is a trigonometric function that might not be immediately integrable in its current form. Therefore, we need to employ some techniques to simplify it. The presence of both sine and cosine functions suggests that trigonometric identities and substitution methods will be key to solving this problem. Before diving into the solution, it's beneficial to understand the behavior of the integrand within the given interval. As $x$ varies from $0$ to $\frac{\pi}{6}$, both $\sin(x)$ and $\cos(x)$ are positive, which means the integrand is also positive. This ensures that the integral represents a positive area. Recognizing such properties beforehand can help in verifying the correctness of our final answer. The challenge lies in transforming the integrand into a form that we can easily integrate. This often involves breaking down complex trigonometric functions into simpler components using identities and then applying a suitable substitution. In the following sections, we will explore the specific steps to achieve this.

Trigonometric Simplification

To effectively evaluate the integral, the first crucial step involves trigonometric simplification. The integrand $ rac{\sin^3(x)}{\cos(x)}$ appears complex, but we can manipulate it using trigonometric identities to make it more manageable. The key identity we'll use here is the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$. We can rewrite $\sin^3(x)$ as $\sin^2(x) \cdot \sin(x)$. Then, using the Pythagorean identity, we substitute $\sin^2(x)$ with $1 - \cos^2(x)$. This transforms our integrand as follows:

sin⁡3(x)cos⁡(x)=sin⁡2(x)⋅sin⁡(x)cos⁡(x)=(1−cos⁡2(x))⋅sin⁡(x)cos⁡(x)\frac{\sin^3(x)}{\cos(x)} = \frac{\sin^2(x) \cdot \sin(x)}{\cos(x)} = \frac{(1 - \cos^2(x)) \cdot \sin(x)}{\cos(x)}

This manipulation is crucial because it introduces a form where we can apply a simple substitution. By expressing the integrand in terms of $\cos(x)$ and $\sin(x)$, we set the stage for a u-substitution, which is a powerful technique for simplifying integrals. The goal here is to transform the integral into a form that is easier to integrate directly. This step of trigonometric simplification is not just about making the integrand look simpler; it's about changing its structure to align with known integration techniques. Without this simplification, the integral would be much harder to tackle. The ability to recognize and apply appropriate trigonometric identities is a fundamental skill in calculus, and this problem serves as a good example of its importance. By rewriting the integrand in this way, we have effectively paved the way for the next step: the u-substitution.

Applying U-Substitution

Following the trigonometric simplification, the next strategic step is applying u-substitution. This technique is particularly useful when we have a composite function within the integrand. In our case, we have expressed the integrand as $ rac{(1 - \cos^2(x)) \cdot \sin(x)}{\cos(x)}$, which suggests a suitable substitution. Let's set $u = \cos(x)$. The derivative of $u$ with respect to $x$ is:

dudx=−sin⁡(x)\frac{du}{dx} = -\sin(x)

Thus, $du = -\sin(x) dx$. Now, we can rewrite the integral in terms of $u$. We also need to change the limits of integration to reflect the new variable. When $x = 0$, $u = \cos(0) = 1$. When $x = \frac{\pi}{6}$, $u = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. The integral now becomes:

∫0π/6(1−cos⁡2(x))⋅sin⁡(x)cos⁡(x)dx=−∫13/21−u2udu\int_0^{\pi / 6} \frac{(1 - \cos^2(x)) \cdot \sin(x)}{\cos(x)} dx = -\int_1^{\sqrt{3} / 2} \frac{1 - u^2}{u} du

Notice the change in the limits of integration and the negative sign that comes from the $du$ substitution. This transformation is crucial because it simplifies the integrand into a rational function, which is generally easier to integrate. The power of u-substitution lies in its ability to convert complex integrals into more manageable forms. By choosing the right substitution, we can often break down seemingly intractable integrals into simpler components. In this case, substituting $u = \cos(x)$ not only simplifies the integrand but also changes the variable of integration, making the integral solvable using basic calculus techniques. The next step involves integrating this new expression with respect to $u$.

Integration and Evaluation

After applying the u-substitution, we now have the integral in a simplified form: $-\int_1^{\sqrt{3} / 2} \frac{1 - u^2}{u} du$. To proceed, we first divide the terms in the numerator by $u$:

1−u2u=1u−u\frac{1 - u^2}{u} = \frac{1}{u} - u

So, the integral becomes:

−∫13/2(1u−u)du-\int_1^{\sqrt{3} / 2} (\frac{1}{u} - u) du

Now, we can integrate term by term. The integral of $\frac{1}{u}$ is $\ln|u|$, and the integral of $u$ is $\frac{u^2}{2}$. Therefore, we have:

−[ln⁡∣u∣−u22]13/2-\left[ \ln|u| - \frac{u^2}{2} \right]_1^{\sqrt{3} / 2}

Next, we evaluate the antiderivative at the limits of integration. We substitute $u = \frac{\sqrt{3}}{2}$ and $u = 1$ into the expression and subtract:

−[(ln⁡(32)−(3/2)22)−(ln⁡(1)−122)]-\left[ \left( \ln(\frac{\sqrt{3}}{2}) - \frac{(\sqrt{3} / 2)^2}{2} \right) - \left( \ln(1) - \frac{1^2}{2} \right) \right]

Simplifying the expression, we get:

−[ln⁡(32)−38−(0−12)]=−ln⁡(32)+38−12=−ln⁡(32)−18-\left[ \ln(\frac{\sqrt{3}}{2}) - \frac{3}{8} - (0 - \frac{1}{2}) \right] = -\ln(\frac{\sqrt{3}}{2}) + \frac{3}{8} - \frac{1}{2} = -\ln(\frac{\sqrt{3}}{2}) - \frac{1}{8}

Since $-\ln(\frac{\sqrt{3}}{2}) = \ln(\frac{2}{\sqrt{3}})$, the final result is:

ln⁡(23)−18\ln(\frac{2}{\sqrt{3}}) - \frac{1}{8}

This is the exact value of the integral. The process of integration and evaluation involves applying the fundamental theorem of calculus, which connects differentiation and integration. By finding the antiderivative and evaluating it at the limits of integration, we obtain the definite integral's value. The simplification of the resulting expression often requires careful attention to detail, especially when dealing with logarithmic and fractional terms. This step is a culmination of all the previous steps, where we transform the integral into a solvable form and then apply the necessary calculus techniques to find the final answer.

Final Answer and Conclusion

After performing trigonometric simplification, applying u-substitution, and integrating, we have arrived at the final answer for the definite integral: $\int_0^{\pi / 6} \frac{\sin ^3(x)}{\cos (x)} d x = \ln(\frac{2}{\sqrt{3}}) - \frac{1}{8}$. This result represents the exact value of the integral, which is the area under the curve of the function $\frac{\sin ^3(x)}{\cos (x)}$ from $x = 0$ to $x = \frac{\pi}{6}$. Throughout this process, we utilized several key calculus techniques, including trigonometric identities and u-substitution, to transform the integral into a manageable form. The success of this evaluation highlights the importance of strategic problem-solving in calculus. Recognizing the structure of the integrand and choosing appropriate techniques are crucial skills for tackling complex integrals. This example serves as a valuable exercise for students and professionals alike, reinforcing the understanding of integral calculus and trigonometric functions. The combination of trigonometric identities and substitution methods is a common theme in calculus problems, and mastering these techniques is essential for success in higher-level mathematics and related fields. In conclusion, the evaluation of this integral demonstrates the power and elegance of calculus in solving seemingly complex problems. By breaking down the problem into smaller, more manageable steps, we were able to arrive at a precise and meaningful answer.