Question Statement

Evaluate the integral:

$\int \frac{dx}{\sqrt{5-x^2}}$

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Background and Explanation

This integral follows the standard form $\int \frac{dx}{\sqrt{a^2-x^2}}$, which is solved using trigonometric substitution. The key insight is recognizing that the denominator contains a difference of squares that suggests the identity $1-{\sin }^2\theta ={\cos }^2\theta$.

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Solution

Let $I$ denote the integral we wish to evaluate:

$I=\int \frac{dx}{\sqrt{5-x^2}}$

First, rewrite the denominator to explicitly identify the constant $a$:

$I=\int \frac{dx}{\sqrt{(\sqrt{5}{)}^2-(x{)}^2}}$

Step 1: Trigonometric Substitution

Since we have the form $\sqrt{a^2-x^2}$ where $a=\sqrt{5}$, we use the substitution:

$x=\sqrt{5}\sin \theta$

Differentiating both sides with respect to $\theta$:

$dx=\sqrt{5}\cos \theta d\theta$

Step 2: Substitute into the Integral

Substituting $x$ and $dx$ into equation (1):

\begin{align*} I &= \int \frac{\sqrt{5} \cos \theta \, d\theta}{\sqrt{(\sqrt{5})^{2}-(\sqrt{5} \sin \theta)^{2}}} \\ &= \int \frac{\sqrt{5} \cos \theta}{\sqrt{5-5\sin^{2}\theta}} \, d\theta \\ &= \int \frac{\sqrt{5} \cos \theta}{\sqrt{5(1-\sin^{2}\theta)}} \, d\theta \end{align*}

Step 3: Simplify Using Trigonometric Identity

Using the identity $1-{\sin }^2\theta ={\cos }^2\theta$:

\begin{align*} I &= \int \frac{\sqrt{5} \cos \theta}{\sqrt{5}\sqrt{\cos^{2}\theta}} \, d\theta \\ &= \int \frac{\sqrt{5} \cos \theta}{\sqrt{5} \cos \theta} \, d\theta \\ &= \int 1 \, d\theta \\ &= \theta + c \end{align*}

Step 4: Back-Substitution

From our original substitution $x=\sqrt{5}\sin \theta$, we solve for $\theta$:

$\sin \theta =\frac{x}{\sqrt{5}}$

Therefore:

$\theta ={\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)$

Substituting this back into equation (2):

$I={\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$

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Key Formulas or Methods Used

• Standard Integral Formula: $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+C$
• Trigonometric Substitution: For $\sqrt{a^2-x^2}$, substitute $x=a\sin \theta$
• Pythagorean Identity: ${\sin }^2\theta +{\cos }^2\theta =1$ (or $1-{\sin }^2\theta ={\cos }^2\theta$)
• Differential Calculation: If $x=a\sin \theta$, then $dx=a\cos \theta d\theta$

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Summary of Steps

1. Identify the form: Rewrite $\sqrt{5-x^2}$ as $\sqrt{(\sqrt{5}{)}^2-x^2}$ to recognize $a=\sqrt{5}$
2. Substitute: Let $x=\sqrt{5}\sin \theta$, which gives $dx=\sqrt{5}\cos \theta d\theta$
3. Transform the integral: Substitute into the original integral to get $\int \frac{\sqrt{5}\cos \theta }{\sqrt{5}\cos \theta }d\theta$
4. Simplify: Cancel terms to obtain $\int 1d\theta =\theta +c$
5. Convert back: From $\sin \theta =\frac{x}{\sqrt{5}}$, get $\theta ={\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)$
6. Final answer: $I={\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)+c$