Question

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

Hint:

For integrals involving inverse trigonometric functions, substitute to simplify the expression and use trigonometric identities to reduce complexity.

Answer 1

We are tasked with solving the given integral step by step. Step 1: Substitution Let: \[ x^{\frac{3}{2}} = t \quad \Rightarrow \quad \frac{3}{2} x^{\frac{1}{2}} dx = dt. \] This simplifies the differential: \[ x^{\frac{1}{2}} dx = \frac{2}{3} dt. \] Step 2: Transform the integral The given integral becomes: \[ \frac{2}{3} \int t \sin^{-1} t \, dt. \] Step 3: Apply integration by parts We use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du. \] Here, choose \(u = \sin^{-1} t\) and \(dv = t \, dt\). Then: \[ du = \frac{1}{\sqrt{1-t^2}} \, dt, \quad v = \frac{t^2}{2}. \] Substituting these into the formula: \[ \frac{2}{3} \int t \sin^{-1} t \, dt = \frac{2}{3} \left[ \sin^{-1} t \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{\sqrt{1-t^2}} \, dt \right]. \] Step 4: Simplify the integral This expands to: \[ \frac{2}{3} \int t \sin^{-1} t \, dt = \frac{2}{3} \left[ \frac{t^2 \sin^{-1} t}{2} - \frac{1}{2} \int \frac{t^2}{\sqrt{1-t^2}} \, dt \right]. \] For the second term, rewrite \(t^2 = (1 - (1-t^2))\), so: \[ \int \frac{t^2}{\sqrt{1-t^2}} \, dt = \int \frac{1 - (1-t^2)}{\sqrt{1-t^2}} \, dt = \int \sqrt{1-t^2} \, dt - \int \frac{1}{\sqrt{1-t^2}} \, dt. \] Step 5: Solve the individual integrals 1. For \(\int \sqrt{1-t^2} \, dt\), the solution is standard: \[ \int \sqrt{1-t^2} \, dt = \frac{1}{2} \left[ t \sqrt{1-t^2} + \sin^{-1} t \right]. \] 2. For \(\int \frac{1}{\sqrt{1-t^2}} \, dt\), the solution is: \[ \int \frac{1}{\sqrt{1-t^2}} \, dt = \sin^{-1} t. \] Substitute these results back into the integral: \[ \int \frac{t^2}{\sqrt{1-t^2}} \, dt = \frac{1}{2} \left[ t \sqrt{1-t^2} + \sin^{-1} t \right] - \sin^{-1} t. \] Simplify: \[ \int \frac{t^2}{\sqrt{1-t^2}} \, dt = \frac{1}{2} t \sqrt{1-t^2} + \frac{1}{2} \sin^{-1} t - \sin^{-1} t. \] \[ \int \frac{t^2}{\sqrt{1-t^2}} \, dt = \frac{1}{2} t \sqrt{1-t^2} - \frac{1}{2} \sin^{-1} t. \] Step 6: Substitute back into the expression The integral becomes: \[ \frac{2}{3} \int t \sin^{-1} t \, dt = \frac{2}{3} \left[ \frac{t^2 \sin^{-1} t}{2} - \frac{1}{2} \left( \frac{1}{2} t \sqrt{1-t^2} - \frac{1}{2} \sin^{-1} t \right) \right]. \] Simplify the terms: \[ \frac{2}{3} \int t \sin^{-1} t \, dt = \frac{2}{3} \left[ \frac{t^2 \sin^{-1} t}{2} - \frac{1}{4} t \sqrt{1-t^2} - \frac{1}{4} \sin^{-1} t \right]. \] Step 7: Back-substitute for \(t\) Recall that \(t = x^{\frac{3}{2}}\). Substituting back: \[ \frac{2}{3} \int t \sin^{-1} t \, dt = \frac{1}{3} \left[ x^3 \sin^{-1}(x^{\frac{3}{2}}) + \frac{x^{\frac{3}{2}}}{2} \sqrt{1 - x^3} - \frac{1}{2} \sin^{-1}(x^{\frac{3}{2}}) \right] + C. \] Final Answer: \[ \boxed{\frac{1}{3} \left[ x^3 \sin^{-1}(x^{\frac{3}{2}}) + \frac{x^{\frac{3}{2}}}{2} \sqrt{1 - x^3} - \frac{1}{2} \sin^{-1}(x^{\frac{3}{2}}) \right] + C.} \]