Indefinite Integration
Integration by Parts
Grade None
Question:
<p>\(\displaystyle\int \sin^{-1}\!\sqrt{\dfrac{x}{a+x}}\,dx\) equals</p>
<li>\(x\sin^{-1}\!\sqrt{\dfrac{x}{a+x}}-\sqrt{ax}+C\)</li>
<li>\((a+x)\sin^{-1}\!\sqrt{\dfrac{x}{a+x}}-\sqrt{ax}+C\)</li>
<li>\((a+x)\tan^{-1}\!\sqrt{\dfrac{x}{a}}-\sqrt{ax}+C\)</li>
<li>\(x\tan^{-1}\!\sqrt{\dfrac{x}{a}}-\sqrt{ax}+C\)</li>
Step-by-Step Solution
Key Concept: Note that sin⁻^1(\sqrt{x/(a+x})) = tan⁻^1(\sqrt{x/a}). Use integration by parts: u=sin⁻^1(\sqrt{x/(a+x})), dv=dx.
<p><strong>Simplify:</strong> Let $\theta = \sin^{-1}\!\sqrt{\dfrac{x}{a+x}}$, so $\sin\theta=\sqrt{\dfrac{x}{a+x}}\Rightarrow\tan\theta=\sqrt{\dfrac{x}{a}}$. Thus the inverse sine equals $\tan^{-1}\!\sqrt{x/a}$.</p>
<p><strong>By parts:</strong> $u=\tan^{-1}\!\sqrt{x/a},\;dv=dx\Rightarrow v=x$.</p>
<p>$$I = x\tan^{-1}\!\sqrt{\frac{x}{a}} - \int\frac{x}{1+x/a}\cdot\frac{1}{2\sqrt{ax}}\,dx$$</p>
<p>$$= x\tan^{-1}\!\sqrt{\frac{x}{a}} - \frac{1}{2\sqrt a}\int\frac{\sqrt x\,dx}{1+x/a}\cdot\ldots = (a+x)\sin^{-1}\!\sqrt{\tfrac{x}{a+x}}-\sqrt{ax}+C$$</p>
<p>Answer: <strong>(B)</strong></p>
Correct Answer: B