Question

Find: \[ \int \frac{\sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right)}{ \sqrt{a + x}} \, dx \]

Hint:

When facing integrals with inverse trigonometric functions, substituting the argument of the inverse function often helps simplify the expression. Remember to differentiate implicitly to handle the change of variables.

Answer 1

Let the given integral be denoted as \(I\): \[ I = \int \frac{\sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right)}{ \sqrt{a + x}} \, dx \] We employ the substitution: \[ u = \sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right) \] This implies: \[ \sin(u) = \frac{x}{\sqrt{a + x}} \] Squaring both sides yields: \[ \sin^2(u) = \frac{x^2}{a + x} \] Differentiating both sides with respect to \(x\): \[ 2 \sin(u) \cos(u) \frac{du}{dx} = \frac{2x}{a + x} - \frac{x^2}{(a + x)^2} \] The next step is to express the integrand in terms of \(u\) and simplify to evaluate the integral.