Trig Substitution: Solving $(5-x^2)^{-3/2}$ Integrals

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Hey everyone! Let's dive into a fascinating integral problem that many students grapple with in their mathematical analysis journey. We're going to explore why the substitution $x = \sqrt{5}\sin(u)$ is often the go-to method for solving indefinite integrals of the form $(5-x^2)^{-3/2}$ over the interval $[-1, 2]$, and why other substitutions might not be as straightforward. This is a crucial concept in calculus, and understanding it deeply will not only help you ace your exams but also give you a more intuitive grasp of integration techniques. So, let's roll up our sleeves and get started!

The Integral Challenge: $(5-x^2)^{-3/2}$

So, you've stumbled upon the integral of $(5-x^2)^{-3/2}$ and you're wondering why the substitution $x = \sqrt{5}\sin(u)$ seems to be the magic bullet. You might be thinking, "Why this particular substitution? Are there other options?" Well, you're asking the right questions! Let's break down why this substitution is so effective and explore why alternatives might lead you down a more complex path. When we're dealing with integrals, especially those involving square roots of the form $a^2 - x^2$, $a^2 + x^2$, or $x^2 - a^2$, trigonometric substitutions often come to the rescue. These substitutions leverage trigonometric identities to simplify the integrand, making it easier to integrate. In our case, we have $(5-x^2)^{-3/2}$, which closely resembles the form $a^2 - x^2$, where $a^2 = 5$. This is a clear signal that a trigonometric substitution might be our best bet. The substitution $x = \sqrt{5}\sin(u)$ is specifically chosen because it allows us to use the Pythagorean identity $\cos^2(u) = 1 - \sin^2(u)$ to simplify the expression under the square root. When we substitute $x = \sqrt{5}\sin(u)$, we get:

$$\sqrt{5 - x^2} = \sqrt{5 - 5\sin^2(u)} = \sqrt{5(1 - \sin^2(u))} = \sqrt{5\cos^2(u)} = \sqrt{5}\cos(u)$$

This transformation is incredibly powerful because it turns a complex square root expression into a simple trigonometric function, $\sqrt{5}\cos(u)$. Now, let's delve into why other substitutions might not be as effective.

Why $x = \sqrt{5}\sin(u)$ Works Wonders

Okay, so why does $x = \sqrt{5}\sin(u)$ work so beautifully? The key lies in how it simplifies the integral. When we use this substitution, we're essentially transforming the integral into a trigonometric form that we can easily handle. Let's walk through the steps to see it in action.

First, we substitute $x = \sqrt{5}\sin(u)$. This means we also need to find $dx$ in terms of $du$. Differentiating $x$ with respect to $u$, we get:

$$\frac{dx}{du} = \sqrt{5}\cos(u)$$

So, $dx = \sqrt{5}\cos(u) du$. Now, let's substitute both $x$ and $dx$ into our integral:

$$\int (5 - x^2)^{-3/2} dx = \int (5 - 5\sin^2(u))^{-3/2} \sqrt{5}\cos(u) du$$

As we saw earlier, $5 - 5\sin^2(u)$ simplifies to $5\cos^2(u)$, so our integral becomes:

$$\int (5\cos^2(u))^{-3/2} \sqrt{5}\cos(u) du = \int 5^{-3/2}(\cos^2(u))^{-3/2} \sqrt{5}\cos(u) du$$

This simplifies further to:

$$\int \frac{1}{5^{3/2}\cos^3(u)} \sqrt{5}\cos(u) du = \int \frac{\sqrt{5}\cos(u)}{5\sqrt{5}\cos^3(u)} du = \int \frac{1}{5\cos^2(u)} du$$

Now we have:

$$\frac{1}{5} \int \sec^2(u) du$$

The integral of $\sec^2(u)$ is simply $\tan(u)$, so we get:

$$\frac{1}{5} \tan(u) + C$$

But wait, we're not done yet! We need to convert back to $x$. Since $x = \sqrt{5}\sin(u)$, we have $\sin(u) = \frac{x}{\sqrt{5}}$. We can use a right triangle to find $\tan(u)$. Imagine a right triangle where the opposite side is $x$ and the hypotenuse is $\sqrt{5}$. The adjacent side would be $\sqrt{5 - x^2}$. Therefore:

$$\tan(u) = \frac{x}{\sqrt{5 - x^2}}$$

So, our final answer is:

$$\frac{1}{5} \frac{x}{\sqrt{5 - x^2}} + C$$

See how smoothly that went? The substitution $x = \sqrt{5}\sin(u)$ turned a seemingly complex integral into a straightforward trigonometric integral. Now, let's explore why other substitutions might not be as effective.

The Pitfalls of Alternative Substitutions

You might be thinking, "Okay, $x = \sqrt{5}\sin(u)$ works, but what if I tried something else?" Let's consider a few alternative substitutions and see why they might not be as helpful. The key issue with other substitutions often boils down to the complexity they introduce into the integral. The beauty of $x = \sqrt{5}\sin(u)$ is that it directly leverages the Pythagorean identity to simplify the square root term. Other substitutions might not have this simplifying effect, leading to more complicated expressions that are difficult to integrate.

1. Algebraic Substitutions: $u = 5 - x^2$

One common approach might be to try an algebraic substitution like $u = 5 - x^2$. Let's see what happens. If $u = 5 - x^2$, then $du = -2x dx$, and $dx = \frac{du}{-2x}$. Substituting this into our integral, we get:

$$\int (5 - x^2)^{-3/2} dx = \int u^{-3/2} \frac{du}{-2x}$$

Now we have a problem: we still have an $x$ in our integral. We can express $x$ in terms of $u$ as $x = \sqrt{5 - u}$, but this leads to a more complicated integral:

$$-\frac{1}{2} \int u^{-3/2} \frac{1}{\sqrt{5 - u}} du$$

This integral is not easily solvable using elementary functions. While it might be possible to solve it using more advanced techniques, it's significantly more complex than the trigonometric substitution.

2. Hyperbolic Substitutions: $x = \sqrt{5}\tanh(u)$

Another option might be to consider hyperbolic substitutions. For instance, we could try $x = \sqrt{5}\tanh(u)$. In this case, $dx = \sqrt{5}\text{sech}^2(u) du$. Substituting this into our integral, we get:

$$\int (5 - x^2)^{-3/2} dx = \int (5 - 5\tanh^2(u))^{-3/2} \sqrt{5}\text{sech}^2(u) du$$

Using the hyperbolic identity $1 - \tanh^2(u) = \text{sech}^2(u)$, we have:

$$\int (5\text{sech}^2(u))^{-3/2} \sqrt{5}\text{sech}^2(u) du = \int \frac{1}{5^{3/2}\text{sech}^3(u)} \sqrt{5}\text{sech}^2(u) du$$

This simplifies to:

$$\frac{1}{5} \int \frac{1}{\text{sech}(u)} du = \frac{1}{5} \int \cosh(u) du = \frac{1}{5} \sinh(u) + C$$

Now we need to convert back to $x$. Since $x = \sqrt{5}\tanh(u)$, we have $\tanh(u) = \frac{x}{\sqrt{5}}$. Using the relationship between hyperbolic functions, we can find $\sinh(u)$, but this process is more involved than the trigonometric substitution. While this substitution does lead to a solution, it's not as straightforward as using $x = \sqrt{5}\sin(u)$.

3. Other Trigonometric Substitutions: $x = \sqrt{5}\cos(u)$

You might also wonder, "What if I used $x = \sqrt{5}\cos(u)$ instead of $x = \sqrt{5}\sin(u)$?" Let's explore this briefly. If $x = \sqrt{5}\cos(u)$, then $dx = -\sqrt{5}\sin(u) du$. Substituting this into our integral, we get:

$$\int (5 - x^2)^{-3/2} dx = \int (5 - 5\cos^2(u))^{-3/2} (-\sqrt{5}\sin(u)) du$$

Using the identity $\sin^2(u) = 1 - \cos^2(u)$, we have:

$$\int (5\sin^2(u))^{-3/2} (-\sqrt{5}\sin(u)) du = \int \frac{1}{5^{3/2}\sin^3(u)} (-\sqrt{5}\sin(u)) du$$

This simplifies to:

$$-\frac{1}{5} \int \frac{1}{\sin^2(u)} du = -\frac{1}{5} \int \csc^2(u) du = \frac{1}{5} \cot(u) + C$$

This substitution works, but converting back to $x$ involves finding $\cot(u)$ in terms of $x$, which is a bit more cumbersome than finding $\tan(u)$ when using $x = \sqrt{5}\sin(u)$.

The Takeaway: Choosing the Right Substitution

So, what's the bottom line? While there might be multiple ways to tackle an integral, some methods are clearly more efficient than others. In the case of $(5-x^2)^{-3/2}$, the substitution $x = \sqrt{5}\sin(u)$ is the hero because it directly simplifies the integral using the Pythagorean identity. Other substitutions, like algebraic or hyperbolic substitutions, might lead to more complicated expressions or require more steps to convert back to the original variable. When you're faced with integrals involving square roots of the form $a^2 - x^2$, $a^2 + x^2$, or $x^2 - a^2$, remember the power of trigonometric substitutions. They're often your best friend in the world of calculus! Keep practicing, and you'll become a master of integration in no time. You've got this!

Definite Integrals and the Interval $[-1, 2]$

Now, let’s briefly touch on the interval $[-1, 2]$ mentioned in the original question. When dealing with definite integrals, we need to consider the limits of integration. If we were to evaluate the definite integral of $(5-x^2)^{-3/2}$ over the interval $[-1, 2]$, we would need to change the limits of integration to correspond to the new variable $u$ after the substitution. Given $x = \sqrt{5}\sin(u)$, we can find the corresponding values of $u$ for $x = -1$ and $x = 2$. However, it’s crucial to ensure that the function is continuous and well-defined over the interval of integration. In this case, $(5-x^2)^{-3/2}$ is continuous over $[-1, 2]$, so we can proceed with the integration.

To find the new limits, we solve for $u$:

$$u = \arcsin(\frac{x}{\sqrt{5}})$$

For $x = -1$, $u_1 = \arcsin(-\frac{1}{\sqrt{5}})$. For $x = 2$, $u_2 = \arcsin(\frac{2}{\sqrt{5}})$. The definite integral would then be evaluated as:

$$\int_{-1}^{2} (5-x^2)^{-3/2} dx = \frac{1}{5} [\frac{x}{\sqrt{5-x^2}}]_{-1}^{2} = \frac{1}{5} (\frac{2}{\sqrt{5-4}} - \frac{-1}{\sqrt{5-1}}) = \frac{1}{5} (2 + \frac{1}{2}) = \frac{1}{5} (\frac{5}{2}) = \frac{1}{2}$$

Understanding the interplay between indefinite integrals, substitutions, and definite integrals is essential for mastering calculus. Keep exploring and keep questioning – that’s the spirit of mathematical inquiry!