Question

Completing the square Evaluate the following integrals. $$\int \frac{d u}{2 u^{2}-12 u+36}$$

Short answer

Question: Evaluate the integral $$\int \frac{d u}{2 u^{2}-12 u+36}$$ Answer: $$\frac{1}{3} \arctan \frac{u - 3}{3} + C$$

Step-by-step solution

Completing the square

First, let's complete the square for the quadratic function in the denominator: $$2u^2 - 12u + 36$$ Divide the quadratic function by 2 to get: $$u^2 - 6u + 18$$ Now, complete the square by adding and subtracting $$\left(\frac{1}{2} \cdot 6\right)^2 = 9$$: $$\begin{aligned} u^2 - 6u + 18 &= (u^2 - 6u + 9) + 18 - 9\\ &= (u - 3)^2 + 9 \end{aligned}$$ Now, the integral becomes: $$\int \frac{d u}{(u - 3)^2 + 9}$$

Substitution

Let's make the substitution: $$\begin{cases} v = u - 3 \\ d v = d u \end{cases}$$ Thus, the integral becomes: $$\int \frac{d v}{v^2 + 9}$$

Recognize the arctangent integral

This integral is now in the form of an inverse tangent (arctangent) integral. The general form is as follows: $$\int \frac{d x}{x^2 + a^2} = \frac{1}{a}\arctan \frac{x}{a} + C$$

Evaluate the integral

Using the arctangent integral formula with \(x=v\) and \(a=3\), we get: $$\int \frac{d v}{v^2 + 9} = \frac{1}{3} \arctan \frac{v}{3} + C$$

Back-substitute

Finally, we back-substitute back to our original variable, \(u\): $$\frac{1}{3} \arctan \frac{v}{3} + C = \frac{1}{3} \arctan \frac{u - 3}{3} + C$$ Thus, the final solution for the integral is: $$\int \frac{d u}{2 u^{2}-12 u+36} = \frac{1}{3} \arctan \frac{u - 3}{3} + C$$

Key concepts

Quadratic Function

A quadratic function is a polynomial function of degree two. It's generally represented in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our problem, we started with the equation \( 2u^2 - 12u + 36 \). One powerful tool for working with quadratic functions is called 'completing the square'.

This technique helps simplify the quadratic into a recognizable form that can be used for further calculus processes.

• First, you divide the leading coefficient if it’s not 1. This makes the quadratic easier to manage.
• Second, split the linear term (\( b \)) by taking half, then squaring it, and adding it into your function inside the brackets.
• Finally, adjust the constant term appropriately outside the brackets.

In our exercise, we divided by 2 to simplify, giving \( u^2 - 6u + 18 \), then used completing the square to get \( (u-3)^2 + 9 \). This organized form prepares it for integral calculus.

Integral Calculus

Integral calculus deals with the accumulation of quantities and the areas under and between curves. One key application is solving definite and indefinite integrals. Our task involves evaluating an indefinite integral, which determines the antiderivative or the original function that was differentiated to yield the integrand.

Using calculus, you can often simplify complex expressions into forms that are easier to integrate. Completing the square is one such method, and it sets the stage for applying specific integral formulas.

• The integral we want to solve, \( \int \frac{du}{(u-3)^2 + 9} \), is an expression ready for substitution or formula recognition.
• Integral calculus includes techniques like substitution and recognizing standard forms like the arctangent integral, which we’ll discuss next.

Solving integrals often involves converting them into a form where known calculus techniques can be applied, which is essential in this exercise.

Arctangent Integral

The arctangent integral is one of the many standard integrals in calculus and relates specifically to inverse trigonometric functions. It is often recognized in the form \( \int \frac{dx}{x^2 + a^2} \), resulting in an expression involving \( \arctan \).

This form is handy as it's directly linked to the inverse tangent function, making it a perfect fit for our integral after substitution.

• The specific formula is \( \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\arctan \frac{x}{a} + C \), where \( a \) is a constant.
• In our case, we compared \( v^2 + 9 \) so identified \( a \) as 3.
• Applying this formula transformed our integral into \( \frac{1}{3} \arctan \frac{v}{3} + C \).

Recognizing the pattern in the form of the arctangent integral simplifies the problem significantly, demonstrating how preset calculus tools streamline solving integrals.

Substitution Method

The substitution method in calculus is an important technique that makes complex integrals simpler. It involves changing variables to rewrite the integral in an easier form.

In our exercise, completing the square set up the problem for a straightforward substitution. We used \( v = u - 3 \) to simplify the integral.

• Once substituted, the integral \( \int \frac{du}{(u-3)^2 + 9} \) became \( \int \frac{dv}{v^2 + 9} \).
• This makes solving the integral easier by reducing it to a standard form that can be integrated using the arctangent formula.
• This method also offers a way to switch back to the original variable, ensuring the final answer is in terms of the initial variable we started with.

Substitution is particularly useful when you can identify a new variable that simplifies the integral into a known or easier-to-integrate form. It nourishes our understanding of transformation and manipulation in calculus.