Question Statement

Evaluate the integral:

$\int \frac{d x}{20 - x ^{2} - 4 x}$

Background and Explanation

This integral contains a quadratic expression under a square root in the denominator. To solve it, we first complete the square to transform the expression into the standard form $a^{2} - u^{2}$ , which allows us to use either trigonometric substitution or the inverse sine integral formula.

Solution

Step 1: Complete the Square

We begin by rewriting the expression under the square root. The denominator is $20 - x^{2} - 4 x$ . To complete the square for the quadratic part $- x^{2} - 4 x$ , we factor out the negative sign and focus on $x^{2} + 4 x$ :

$\begin{align*} I &= \int \frac{dx}{\sqrt{20-x^{2}-4 x}} \\ &= \int \frac{1}{\sqrt{20+4-4-x^{2}-4 x}} dx \quad \text{(adding and subtracting 4)} \\ &= \int \frac{1}{\sqrt{24-\left(x^{2}+4 x+4\right)}} dx \\ &= \int \frac{1}{\sqrt{(\sqrt{24})^{2}-(x+2)^{2}}} dx \end{align*}$

Here we used the completion of square: $x^{2} + 4 x + 4 = (x + 2)^{2}$ , and noted that $20 + 4 = 24 = (24)^{2}$ .

Step 2: Trigonometric Substitution

The integral is now in the form $\int \frac{d u}{a ^{2} - u ^{2}}$ where $a = 24$ and $u = x + 2$ . We use the substitution:

$x + 2 = 24 sin θ$

Taking differentials: $d x = 24 cos θ d θ$

Step 3: Substitute and Simplify

Substituting into equation (1):

$\begin{align*} I &= \int \frac{1}{\sqrt{(\sqrt{24})^{2}-(\sqrt{24} \sin \theta)^{2}}} \cdot \sqrt{24} \cos \theta \, d\theta \\ &= \int \frac{\sqrt{24} \cos \theta \, d\theta}{\sqrt{24-24 \sin ^{2} \theta}} \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24\left(1-\sin ^{2} \theta\right)}} d\theta \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24 \cos ^{2} \theta}} d\theta \quad \text{(using } 1-\sin^2\theta = \cos^2\theta\text{)} \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24} \cos \theta} d\theta \\ &= \int 1 \, d\theta \\ &= \theta + c \end{align*}$

Step 4: Back-Substitute to Express in Terms of $x$

From our substitution: $x + 2 sin θ θ = 24 sin θ = \frac{x + 2}{24} = sin^{- 1} (\frac{x + 2}{24})$

Substituting this back into equation (2):

$I = sin^{- 1} (\frac{x + 2}{24}) + c$

This can also be written as: $I = sin^{- 1} (\frac{x + 2}{2 6}) + c$

Key Formulas or Methods Used

Completing the square: $x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}$
Standard integral form: $\int \frac{d u}{a ^{2} - u ^{2}} = sin^{- 1} (\frac{u}{a}) + C$
Trigonometric substitution: For $a^{2} - u^{2}$ , substitute $u = a sin θ$
Pythagorean identity: $1 - sin^{2} θ = cos^{2} θ$
Differential: If $u = a sin θ$ , then $d u = a cos θ d θ$

Summary of Steps

Complete the square in the denominator to rewrite $20 - x^{2} - 4 x$ as $24 - (x + 2)^{2}$
Identify the form $a^{2} - u^{2}$ with $a = 24$ and $u = x + 2$
Substitute $x + 2 = 24 sin θ$ and $d x = 24 cos θ d θ$
Simplify the integrand using the identity $1 - sin^{2} θ = cos^{2} θ$ , which causes the square root to cancel with the numerator
Integrate $\int 1 d θ = θ + c$
Back-substitute using $θ = sin^{- 1} (\frac{x + 2}{24})$ to express the answer in terms of $x$

Step 1: Complete the Square

We begin by rewriting the expression under the square root. The denominator is

20 - x^{2} - 4 x

. To complete the square for the quadratic part

- x^{2} - 4 x

, we factor out the negative sign and focus on

x^{2} + 4 x

\begin{align*} I &= \int \frac{dx}{\sqrt{20-x^{2}-4 x}} \\ &= \int \frac{1}{\sqrt{20+4-4-x^{2}-4 x}} dx \quad \text{(adding and subtracting 4)} \\ &= \int \frac{1}{\sqrt{24-\left(x^{2}+4 x+4\right)}} dx \\ &= \int \frac{1}{\sqrt{(\sqrt{24})^{2}-(x+2)^{2}}} dx \end{align*}

Here we used the completion of square:

x^{2} + 4 x + 4 = (x + 2)^{2}

, and noted that

20 + 4 = 24 = (24)^{2}

Step 3: Substitute and Simplify

Substituting into equation (1):

\begin{align*} I &= \int \frac{1}{\sqrt{(\sqrt{24})^{2}-(\sqrt{24} \sin \theta)^{2}}} \cdot \sqrt{24} \cos \theta \, d\theta \\ &= \int \frac{\sqrt{24} \cos \theta \, d\theta}{\sqrt{24-24 \sin ^{2} \theta}} \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24\left(1-\sin ^{2} \theta\right)}} d\theta \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24 \cos ^{2} \theta}} d\theta \quad \text{(using } 1-\sin^2\theta = \cos^2\theta\text{)} \\ &= \int \frac{\sqrt{24} \cos \theta}{\sqrt{24} \cos \theta} d\theta \\ &= \int 1 \, d\theta \\ &= \theta + c \end{align*}

Summary of Steps

Complete the square in the denominator to rewrite

20 - x^{2} - 4 x

24 - (x + 2)^{2}

Identify the form

a^{2} - u^{2}

with

a = 24

and

u = x + 2

Substitute

x + 2 = 24 sin θ

and

d x = 24 cos θ d θ

Simplify the integrand using the identity

1 - sin^{2} θ = cos^{2} θ

, which causes the square root to cancel with the numerator

Integrate

\int 1 d θ = θ + c

Back-substitute using

θ = sin^{- 1} (\frac{x + 2}{24})

to express the answer in terms of

x

3.3 Q-6

Question Statement

Background and Explanation

Solution

Step 1: Complete the Square

Step 2: Trigonometric Substitution

Step 3: Substitute and Simplify

Step 4: Back-Substitute to Express in Terms of $x$

Key Formulas or Methods Used

Summary of Steps

3.3 Q-6

Question Statement

Background and Explanation

Solution

Step 1: Complete the Square

Step 2: Trigonometric Substitution

Step 3: Substitute and Simplify

Step 4: Back-Substitute to Express in Terms of $x$

Key Formulas or Methods Used

Summary of Steps