Definite Integration
Integration by substitution
Grade Class 12
Question:
$\int \sin^{-1} \sqrt{\frac{x}{a+x}} dx$ equals
(A) $(a+x)\arctan\sqrt{\frac{x}{a}} - \sqrt{ax} + C$
(B) $(a+x)\arctan\sqrt{\frac{a}{x}} - \sqrt{ax} + C$
(C) $(a+x^2)\arctan\sqrt{\frac{x^3}{a}} + \sqrt{ax} + C$
(D) $(a+x)\arctan\sqrt{x} + \sqrt{\frac{a}{x}} + C$
Step-by-Step Solution
Key Concept: Use substitution x = a tan^2(theta) to simplify the integrand.
Let $x = a \tan^2 \theta$, then $dx = 2a \tan \theta \sec^2 \theta d\theta$. The integral becomes $\int \sin^{-1}(\sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}}) 2a \tan \theta \sec^2 \theta d\theta = \int \sin^{-1}(\sin \theta) 2a \tan \theta \sec^2 \theta d\theta = \int \theta (2a \tan \theta \sec^2 \theta) d\theta$. Using integration by parts, we get $\theta (a \tan^2 \theta) - \int a \tan^2 \theta d\theta = a \theta \tan^2 \theta - a \int (\sec^2 \theta - 1) d\theta = a \theta \tan^2 \theta - a \tan \theta + a \theta + C = a \theta (1 + \tan^2 \theta) - a \tan \theta + C = a \theta \sec^2 \theta - a \tan \theta + C$. Substituting back $\theta = \tan^{-1} \sqrt{x/a}$, we get $(a+x) \tan^{-1} \sqrt{x/a} - \sqrt{ax} + C$.
Correct Answer: A