Problem 73 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{x+1}{\sqrt{10 x-x^{2}}} d x $$

Step-by-Step Solution

Answer

\( \int \frac{x+1}{\sqrt{10x-x^2}} \, dx = -5\sqrt{1 - \left(\frac{x-5}{5}\right)^2} + 6\sin^{-1}\left(\frac{x-5}{5}\right) + C \)

Step 1: Identify the Type of Integral

The integral \( \int \frac{x+1}{\sqrt{10x-x^2}} \, dx \) appears to be a rational function divided by a square root of a quadratic expression. This suggests it might involve a trigonometric substitution or a complete the square method.

Step 2: Complete the Square

To simplify the integral, transform the quadratic expression under the square root. The expression \(10x - x^2\) can be rewritten as \(-(x^2 - 10x)\). Completing the square gives \(x^2 - 10x = (x-5)^2 - 25\). Therefore, \(-(x^2 - 10x) = 25 - (x-5)^2\), and so \(\sqrt{10x - x^2} = \sqrt{25 - (x-5)^2}\).

Step 3: Use Trigonometric Substitution

With \(\sqrt{25 - (x-5)^2}\), we can use the trigonometric substitution \(x - 5 = 5\sin\theta\). Therefore, \(dx = 5\cos\theta\, d\theta\) and \(\sqrt{25 - (x-5)^2} = 5\cos\theta\). Substituting, the integral becomes \(\int \frac{5\sin\theta + 6}{5\cos\theta} \cdot 5\cos\theta \, d\theta\).

Step 4: Simplify the Integral

This simplifies to \(\int (5\sin\theta + 6) \, d\theta\). Split this into two separate integrals: \(5\int \sin\theta \, d\theta + 6\int 1 \, d\theta\).

Step 5: Integrate Each Term

The integral \(\int \sin\theta \, d\theta = -\cos\theta\) and \(\int 1 \, d\theta = \theta\). So the integral becomes \(-5\cos\theta + 6\theta + C\), where \(C\) is the constant of integration.

Step 6: Back Substitution

Recall that \(x - 5 = 5\sin\theta\), so \(\sin\theta = \frac{x-5}{5}\) and \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{x-5}{5}\right)^2}\). Hence, \(\theta = \sin^{-1}\left(\frac{x-5}{5}\right)\). Substitute back, giving the final expression: \(-5\sqrt{1 - \left(\frac{x-5}{5}\right)^2} + 6\sin^{-1}\left(\frac{x-5}{5}\right) + C\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a handy technique in integral calculus for simplifying integrals that involve square roots of certain quadratic expressions. When you see an integral involving forms like \(\sqrt{a^2 - x^2}\), a trigonometric substitution can convert it into a simpler expression. In the given exercise, we used this method to handle the square root, \(\sqrt{25 - (x-5)^2}\). Here, we set \(x - 5 = 5\sin\theta\). This substitution is chosen because it transforms the expression under the square root into \(5\cos\theta\), making the integral much easier to solve.

Key steps for using trigonometric substitution include:

• Identifying the right trigonometric substitution based on the form of the quadratic.
• Transforming the original integral into an integral that uses trigonometric functions.
• Solving the new integral with basic trigonometric integration rules.
• Substituting back the original expression at the end using inverse trigonometric functions to return in terms of \(x\).

This technique leverages the Pythagorean identity and can often turn a complicated algebraic problem into a straightforward trigonometric one.

Completing the Square

Completing the square is a powerful algebraic strategy used to transform a quadratic expression into a perfect square binomial. In this exercise, we applied this technique to the quadratic expression under the square root, \(-x^2 + 10x\). Converting it into \((x-5)^2 - 25\) allowed us to see the expression in terms that fit the Pythagorean identity form, helping us prepare for the subsequent trigonometric substitution.

The steps to complete the square include:

• Taking the quadratic expression \(ax^2 + bx + c\).
• Rewriting it in the form \((x-h)^2 + k\), where \(h\) and \(k\) are constants.
• This allows us to identify terms easily manageable by trigonometric substitution.

The benefit of this method is it transforms complex-looking expressions into an easier format that fits well with other integration techniques.

Rational Functions

A rational function is simply a ratio of two polynomials, much like the function \(\frac{x+1}{\sqrt{10x - x^2}}\) encountered in this problem. Rational functions often appear in integrals where different techniques may be needed due to the degree or the complexity of the polynomial involved. In this exercise, the presence of a square root in the denominator made straightforward algebraic integration challenging, thus requiring more advanced techniques like completing the square and trigonometric substitution.

Key points when dealing with rational functions in integrals:

• Understand the degree of the polynomials in the numerator and denominator.
• Simplify the function as much as possible before attempting substitution.
• Identify if algebraic manipulation can convert the function to a simpler one that enables trigonometric or other advanced integration techniques.
• Navigating rational functions frequently opens avenues for partial fraction decomposition in other scenarios, depending on the structure.

Integration Techniques

Integration techniques are a diverse set of methods used to find antiderivatives or integrals of various types of functions. Some common techniques used include substitution, integration by parts, and specific methods for rational, trigonometric, and logarithmic functions. The original exercise involved a combination of completing the square and trigonometric substitution, showcasing how often techniques are used in unison.

Here are a few pivotal integration techniques often employed:

• **Direct integration**, useful for simple, standard functions.
• **Trigonometric substitution**, applied to integrals involving square roots of quadratic expressions.
• **Partial fraction decomposition**, generally used for rational functions when the denominator can be factored into linear terms.
• **Integration by parts**, useful when the integrand is a product of two functions that can be differentiated and integrated.

Understanding when and how to apply these techniques efficiently is crucial. Often integrals require a blend of methods, offering a delightful challenge that deepens algebraic and analytic skills.