Question

Evaluate the integral \( \int \frac{\sin^{-1} \left(\frac{x}{\sqrt{a + x}}\right) \sqrt{a + x}}{dx} \):

• A. \((a + x) \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) - \sqrt{ax} + C\)
• B. \(\frac{1}{a + x} \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) - \sqrt{ax} + C\)
• C. \((a + x) \tan^{-1}\left(\frac{a}{x}\right) + \sqrt{ax} + C\)
• D. \(\sqrt{a} + x \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) + ax + C\)

Hint:

For integrals involving inverse trigonometric functions, consider using substitutions that simplify the argument of the function, especially when it involves square roots.

Answer 1

The Correct Option is A

To solve \( \int \frac{\sin^{-1} \left(\frac{x}{\sqrt{a + x}}\right)}{\sqrt{a + x}} \, dx \), use the substitution \( u = \frac{x}{\sqrt{a + x}} \). 
Then, \( du = \frac{dx}{\sqrt{a + x}} - \frac{x dx}{2(a + x)^{3/2}} \), simplifying to \( dx = \sqrt{a + x} \, du \) after some algebraic manipulation. Substitute and solve: \[ \int \sin^{-1}(u) \, du = u \sin^{-1}(u) + \sqrt{1 - u^2} + C \] Substituting back for \( x \) and simplifying yields: \[ (a + x) \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) - \sqrt{ax} + C \]