Calculation integral of x^2 / sqrt (16 - x^2) dx

Introduction

In the realm of mathematics, integration plays a pivotal role in computing various quantities, such as areas, volumes, and accumulated changes, that arise in diverse scientific and engineering disciplines. One interesting problem is evaluating the integral of $\frac{{�}^2}{\sqrt{16-{�}^2}}��$. In this article, we will delve into the step-by-step process of solving this integral, showcasing the power of trigonometric substitution.

Part 1: The Integral Challenge

The integral in question is:

$$\int \frac{{�}^2}{\sqrt{16-{�}^2}}��$$

It may appear daunting at first glance, but with a clever choice of substitution, we can simplify it and make it amenable to integration.

Part 2: The Trigonometric Substitution

To tackle this integral, we'll perform a trigonometric substitution. First, observe that the denominator $\sqrt{16-{�}^2}$ bears a resemblance to the Pythagorean trigonometric identity:

$${\sin }^2(�)+{\cos }^2(�)=1$$

Let's make the substitution:

$$�=4\sin (�)$$

Differentiating both sides with respect to $�$, we get:

$$��=4\cos (�)��$$

Now, we can express ${�}^2$ in terms of $\sin (�)$:

$${�}^2=(4\sin (�){)}^2=16{\sin }^2(�)$$

Substituting these expressions back into the integral, we obtain:

$$\int \frac{{�}^2}{\sqrt{16-{�}^2}}��=\int \frac{16{\sin }^2(�)}{\sqrt{16-16{\sin }^2(�)}}\cdot 4\cos (�)��$$

Simplifying further:

$$\int \frac{4{\sin }^2(�)}{\sqrt{1-{\sin }^2(�)}}\cdot 4\cos (�)��$$

Part 3: Unraveling the Trigonometric Identity

We're now left with the integral:

$$\int \frac{4{\sin }^2(�)}{\sqrt{1-{\sin }^2(�)}}\cdot 4\cos (�)��$$

To proceed, we employ the Pythagorean identity ${\sin }^2(�)+{\cos }^2(�)=1$ to rewrite ${\sin }^2(�)$ as $1-{\cos }^2(�)$:

$$=\int \frac{4(1-{\cos }^2(�))}{\sqrt{1-(1-{\cos }^2(�))}}\cdot 4\cos (�)��$$

Now, we have:

$$=\int \frac{4(1-{\cos }^2(�))}{\sqrt{{\cos }^2(�)}}\cdot 4\cos (�)��$$

Part 4: Addressing Absolute Value

In the given integral, $�$ can range from $-\frac{�}{2}$ to $\frac{�}{2}$ because of our substitution $�=4\sin (�)$. In this range, $\cos (�)$ is always positive. Therefore, we can simplify the integral as follows:

$$=\int \frac{4(1-{\cos }^2(�))}{\cos (�)}\cdot 4\cos (�)��$$

Now, we are set to integrate term by term:

$$=\int (16-16{\cos }^2(�))��$$

The integral simplifies to:

$$=16�-16\int {\cos }^2(�)��$$

Part 5: Integrating ${\cos }^2(�)$

To integrate ${\cos }^2(�)$, we utilize the half-angle identity ${\cos }^2(�)=\frac{1+\cos (2�)}{2}$:

$$=16�-16(\frac{1}{2}\int 1��+\frac{1}{2}\int \cos (2�)��)$$

Now, we integrate each term:

$$=8�-8\sin (2�)+�$$

where $�$ is the constant of integration.

Part 6: Back to $�$ Domain

We must now revert to the original variable $�$ using the initial trigonometric substitution $�=4\sin (�)$:

$$=8\arcsin \left(\frac{�}{4}\right)-8�+�$$

Part 7: Final Simplification

For the final step, we can simplify further:

$$=8\arcsin \left(\frac{�}{4}\right)-8�+�$$

So, the integral of $\frac{{�}^2}{\sqrt{16-{�}^2}}��$ evaluates to $8\arcsin \left(\frac{�}{4}\right)-8�+�$, where $�$ represents the constant of integration.

Conclusion

In this article, we've navigated through the process of evaluating the integral $\frac{{�}^2}{\sqrt{16-{�}^2}}��$. By employing trigonometric substitution and a series of algebraic and trigonometric identities, we successfully transformed the integral into a manageable form, eventually arriving at a simplified expression. Integration, as demonstrated here, is a powerful mathematical tool with diverse applications in various fields of science and engineering.