Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>\(\displaystyle\int \frac{1}{x^2\sqrt{1-x^2}}\,dx\) equals</p>
<li>\(-\dfrac{\sqrt{1-x^2}}{x}+\ln|x|+C\)</li>
<li>\(-\dfrac{\sqrt{1-x^2}}{x}+C\)</li>
<li>\(\dfrac{\sqrt{1-x^2}}{x}+C\)</li>
<li>\(-\dfrac{\sqrt{1-x^2}}{x}-\sin^{-1}x+C\)</li>
Step-by-Step Solution
Key Concept: Substitute x = sin\theta, or equivalently multiply numerator and denominator to rationalise and use d/dx(\sqrt{1-x^2}/x) = -1/(x^\sqrt[2]{1-x^2}).
<p><strong>Method:</strong> Recognise that $\dfrac{d}{dx}\!\left(\dfrac{\sqrt{1-x^2}}{x}\right) = \dfrac{-1}{x^2\sqrt{1-x^2}}\cdot\ldots$</p>
<p>Let $x=\sin\theta$, $dx=\cos\theta\,d\theta$:</p>
<p>$$\int\frac{\cos\theta\,d\theta}{\sin^2\theta\cdot\cos\theta} = \int\csc^2\theta\,d\theta = -\cot\theta+C$$</p>
<p>Back-substitute: $\cot\theta = \dfrac{\cos\theta}{\sin\theta}=\dfrac{\sqrt{1-x^2}}{x}$.</p>
<p>Hence the answer is $-\dfrac{\sqrt{1-x^2}}{x}+C$. Answer: <strong>(B)</strong></p>
Correct Answer: B