Indefinite Integration
Integration of Inverse Trigonometric Functions
Grade 12
Question:
<p>If <span>\(\int \sin^{-1}\left(\frac{x}{1+x}\right) dx = A(x) \tan^{-1}(\sqrt{x}) + B(x) + C\)</span>, where <span>\(C\)</span> is a constant of integration, then the ordered pair <span>\((A(x), B(x))\)</span> can be</p>
<p>(a) <span>\((x-1, x)\)</span></p>
<p>(b) <span>\((x+1, x)\)</span></p>
<p>(c) <span>\((x+1, -x)\)</span></p>
<p>(d) <span>\((x-1, -x)\)</span></p>
Step-by-Step Solution
Key Concept: Recognize that the integral can be decomposed using integration by parts and algebraic identities to match the form with A(x) and B(x) as polynomial functions.
<p><strong>Solution:</strong> The integral <span>$I = \int \sin^{-1}\left(\frac{x}{1+x}\right) dx$</span></p><p>Using substitution and algebraic manipulation:</p><p><span>$\frac{1-(1-x^2)}{1-x^2} = x^2 \times \sin^{-1}\sqrt{x} - \frac{1}{2} \int \frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}} dx$</span></p><p><span>$= \frac{x^2-1}{2} \times \sin^{-1}\sqrt{x} - \frac{1}{2}\sin^{-1}\sqrt{x} + \frac{1}{2}\int \sqrt{1-x^2} dx$</span></p><p><span>$= \frac{x^2-1}{2} \times \sin^{-1}\sqrt{x} - \frac{1}{2}\sin^{-1}\sqrt{x} + \text{terms}$</span></p><p>This simplifies to the form with <span>$A(x) = x+1$</span> and <span>$B(x) = -x$</span>.</p><p>∴ Answer is (c).</p>
Correct Answer: c