Problem 63 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{x^{2}-8 x+16}{\left(9+8 x-x^{2}\right)^{3 / 2}} d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral $$\int \frac{x^2-8x+16}{(9+8x-x^2)^{(3/2)}}dx$$ using the trigonometric substitution method. Answer: Using the trigonometric substitution method, we found that the integral is equivalent to $$\int \frac{3(3u^2-8u+4)}{(9(1-u^2)+24u)^{3 / 2}} du$$, where \(u = \sin(\theta)\). After evaluating the integral using partial fraction decomposition or another integration technique and substituting back theta with the expression \(\theta = \arcsin\left(\frac{x}{3}\right)\), we can obtain the final result.

Step 1: Identify the trigonometric substitution

Recall that a trigonometric substitution is a method of transforming an integral involving a radical into a simpler form by using identities. In this case, let's use the following substitution: $$x = 3\sin(\theta)$$ Differentiate both sides with respect to theta: $$dx = 3\cos(\theta)d\theta$$ Now we can try to rewrite the integral in terms of theta.

Step 2: Substitute x and dx in the integral

We'll rewrite the given integral with the new substitution. $$\int \frac{(3\sin(\theta))^{2}-8(3\sin(\theta))+16}{\left(9+8(3\sin(\theta))-(3\sin(\theta))^{2}\right)^{3 / 2}} \cdot 3\cos(\theta) d\theta$$

Step 3: Simplify the integral

Simplify the expression inside the integral: $$\int \frac{(9\sin^2(\theta)-24\sin(\theta)+16)}{\left(9(1-\sin^2(\theta))+24\sin(\theta)\right)^{3 / 2}} \cdot 3\cos(\theta) d\theta$$ Now, using the identity \(1-\sin^2(\theta)=\cos^2(\theta)\), we get $$\int \frac{(9\sin^2(\theta)-24\sin(\theta)+16)}{\left(9\cos^2(\theta)+24\sin(\theta)\right)^{3 / 2}} \cdot 3\cos(\theta) d\theta$$

Step 4: Solve the integral

To solve this integral, we can perform a change of variables so that it becomes a rational function, which is easier to integrate. Let's use the substitution: $$u = \sin(\theta)$$ Then, \(du = \cos(\theta) d\theta\). So, the integral becomes: $$\int \frac{(9u^2-24u+16)}{(9(1-u^2)+24u)^{3 / 2}} \cdot 3 du$$ Now, to make it simpler, note that the numerator can be factored: $$\int \frac{3(3u^2-8u+4)}{(9(1-u^2)+24u)^{3 / 2}} du$$ At this point, you can use either partial fraction decomposition or any other integration technique to solve the rest of the integral. In order to get the result in terms of x, substitute back in for u: $$u = \sin(\theta) \Rightarrow \theta = \arcsin\left(\frac{x}{3}\right)$$ Substitute this back into the integral and simplify to obtain the final result.

Key Concepts

Integral Calculus

Integral calculus acts as a powerful tool used to find the accumulated quantities, such as areas under curves or in this case, the solution to a complex integral problem. When dealing with an integral like \( \int \frac{x^{2}-8 x+16}{(9+8 x-x^{2})^{3 / 2}} \, dx \), the main goal is to simplify the integrand to a form that is easier to evaluate.
This often involves transforming the existing expression using substitutions like trigonometric identities. Integral calculus leverages these transformations to not just compute the definite and indefinite integrals but also to tackle various derivatives in reverse, which is crucial in solving physics and engineering problems.
In this context, trigonometric substitution is a brilliant technique to simplify and solve integrals involving radicals and quadratic expressions.

Trigonometric Identities

In calculus, particularly when working with integrals, trigonometric identities are crucial for simplifying expressions that involve radicals. A trigonometric substitution is chosen based on patterns recognizable by these identities:
For instance, when we have a \(\sqrt{a^2 - x^2} \), we might use \( x = a\sin(\theta) \); in the exercise, \( x = 3\sin(\theta) \) is used because it simplifies the under-root expression in terms of a basic trigonometric identity. These identities, such as \( \,1 - \sin^2(\theta) = \cos^2(\theta)\,\), help to convert the integral into a trigonometric form that can be tackled more smoothly.
They transform the integrand into a more manageable form, often allowing the integral to reduce to a standard function that is already known and can easily be evaluated.

Integration Techniques

Integration techniques are essential tools in calculus for addressing complex integrals that cannot be solved by simple antiderivatives. In our problem, after trigonometric substitution is performed, further simplification involves changing variables.
This technique transforms the problem, turning a challenging trigonometric integral into a more straightforward rational function form. Moreover, factoring the polynomial in the numerator, such as \( 9u^2 - 24u + 16 \), enables integration by breaking the integral into simpler parts with techniques like partial fraction decomposition. In cases where substitution methods simplify the integrand significantly, students can replace one set of variables with another to simplify the integration step-by-step. Eventually, substituting back the original variables to derive the solution in terms of \(x\) connects the transformation techniques with the problem's requirements, providing a holistic approach to solving complex integrals.