Question

Find: \[ \int x^2 \sin^{-1}(x^{3/2}) \, dx. \]

Hint:

Use substitution and trigonometric identities effectively for integrals involving inverse trigonometric functions.

Answer 1

Step 1: Use substitution.
Let \( u = \sin^{-1}(x^{3/2}) \). Then: \[ x^{3/2} = \sin u \quad \Rightarrow \quad x = (\sin u)^{2/3}. \] Differentiate: \[ dx = \frac{2}{3} (\sin u)^{-1/3} \cos u \, du. \] Step 2: Substitute into the integral.
Replace \( x \) and \( dx \) in the integral: \[ \int x^2 \sin^{-1}(x^{3/2}) \, dx = \int \left[(\sin u)^{4/3}\right] u \cdot \frac{2}{3} (\sin u)^{-1/3} \cos u \, du. \] Simplify: \[ \int x^2 \sin^{-1}(x^{3/2}) \, dx = \frac{2}{3} \int u (\sin u)^{3/3} \cos u \, du = \frac{2}{3} \int u \sin u \cos u \, du. \] Step 3: Simplify using trigonometric identities.
Use \( \sin u \cos u = \frac{1}{2} \sin(2u) \): \[ \frac{2}{3} \int u \sin u \cos u \, du = \frac{1}{3} \int u \sin(2u) \, du. \] Step 4: Solve using integration by parts.
Let \( v = u \) and \( dw = \sin(2u) \, du \): \[ v = u, \quad dv = du, \quad w = -\frac{1}{2} \cos(2u). \] Integration by parts: \[ \int u \sin(2u) \, du = -\frac{u}{2} \cos(2u) + \frac{1}{2} \int \cos(2u) \, du. \] Step 5: Integrate the remaining term.
\[ \int \cos(2u) \, du = \frac{1}{2} \sin(2u). \] Thus: \[ \int u \sin(2u) \, du = -\frac{u}{2} \cos(2u) + \frac{1}{4} \sin(2u). \] Step 6: Back-substitute \( u = \sin^{-1}(x^{3/2}) \).
Replace \( u \) and simplify to express the final answer. Conclusion:
The value of the integral is: \[ \boxed{-\frac{u}{2} \cos(2u) + \frac{1}{4} \sin(2u) + C}. \]