Questions: Question 9 12 pts Consider the definite integral ∫ f(x)/√g(x) dx where f(x)=x and g(x)=15-2x-x^2. a. Rewrite √g(x) in the form of √(b^2-(x+a)^2). b. Use a trigonometric substitution to evaluate the indefinite integral. Be sure to draw the right triangle and express your final answer in terms of x as much as possible. You must show all steps.

Question 9
12 pts

Consider the definite integral
∫ f(x)/√g(x) dx where f(x)=x and g(x)=15-2x-x^2.
a. Rewrite √g(x) in the form of √(b^2-(x+a)^2).
b. Use a trigonometric substitution to evaluate the indefinite integral. Be sure to draw the right triangle and express your final answer in terms of x as much as possible.
You must show all steps.

Transcript text: Question 9 12 pts Consider the definite integral $\int \frac{f(x)}{\sqrt{g(x)}} d x$ where $f(x)=x$ and $g(x)=15-2 x-x^{2}$. a. Rewrite $\sqrt{g(x)}$ in the form of $\sqrt{b^{2}-(x+a)^{2}}$. b. Use a trigonometric substitution to evaluate the indefinite integral. Be sure to draw the right triangle and express your final answer in terms of $x$ as much as possible. You must show all steps.

Solution

Solution Steps

Solution Approach

a. To rewrite $\sqrt{g(x)}$ in the form $\sqrt{b^2 - (x+a)^2}$, we need to complete the square for the quadratic expression $15 - 2x - x^2$.

b. For the trigonometric substitution, we can use $x + a = b \sin(\theta)$ where $b$ and $a$ are constants determined from the completed square form. Then, we substitute and integrate using trigonometric identities and finally convert back to $x$.

Step 1: Rewrite $\sqrt{g(x)}$ in the form of $\sqrt{b^{2}-(x+a)^{2}}$

Given $ g(x) = 15 - 2x - x^2 $, we need to rewrite it in the form $\sqrt{b^2 - (x + a)^2}$.

First, complete the square for $ g(x) $:

\[ g(x) = 15 - 2x - x^2 = -\left(x^2 + 2x - 15\right) \]

Complete the square inside the parentheses:

\[ x^2 + 2x - 15 = (x^2 + 2x + 1) - 1 - 15 = (x + 1)^2 - 16 \]

Thus,

\[ g(x) = -\left((x + 1)^2 - 16\right) = 16 - (x + 1)^2 \]

So,

\[ \sqrt{g(x)} = \sqrt{16 - (x + 1)^2} \]

Step 2: Use a trigonometric substitution to evaluate the indefinite integral

We use the trigonometric substitution $ x + 1 = 4 \sin \theta $. Then,

\[ dx = 4 \cos \theta \, d\theta \]

Substitute $ x + 1 = 4 \sin \theta $ into the integral:

\[ \int \frac{x}{\sqrt{16 - (x + 1)^2}} \, dx = \int \frac{4 \sin \theta - 1}{\sqrt{16 - 16 \sin^2 \theta}} \cdot 4 \cos \theta \, d\theta \]

Simplify the square root:

\[ \sqrt{16 - 16 \sin^2 \theta} = \sqrt{16 (1 - \sin^2 \theta)} = \sqrt{16 \cos^2 \theta} = 4 \cos \theta \]

The integral becomes:

\[ \int \frac{4 \sin \theta - 1}{4 \cos \theta} \cdot 4 \cos \theta \, d\theta = \int (4 \sin \theta - 1) \, d\theta \]

Separate the integral:

\[ \int 4 \sin \theta \, d\theta - \int 1 \, d\theta \]

Evaluate each part:

\[ \int 4 \sin \theta \, d\theta = -4 \cos \theta \]

\[ \int 1 \, d\theta = \theta \]

Thus, the integral is:

\[ -4 \cos \theta - \theta + C \]

Step 3: Express the final answer in terms of $ x $

Recall the substitution $ x + 1 = 4 \sin \theta $:

\[ \sin \theta = \frac{x + 1}{4} \]

Using the right triangle, we have:

\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{x + 1}{4}\right)^2} = \sqrt{\frac{16 - (x + 1)^2}{16}} = \frac{\sqrt{16 - (x + 1)^2}}{4} \]

So,

\[ \cos \theta = \frac{\sqrt{16 - (x + 1)^2}}{4} \]

And,

\[ \theta = \arcsin \left(\frac{x + 1}{4}\right) \]

Substitute back into the integral:

\[ -4 \cos \theta - \theta + C = -4 \left(\frac{\sqrt{16 - (x + 1)^2}}{4}\right) - \arcsin \left(\frac{x + 1}{4}\right) + C \]

Simplify:

\[ -\sqrt{16 - (x + 1)^2} - \arcsin \left(\frac{x + 1}{4}\right) + C \]