Question Statement

Evaluate the integral:

$\int x^2{\sin }^{-1}xdx$

---

Background and Explanation

This problem requires integration by parts (since we have a product of an algebraic function $x^2$ and an inverse trigonometric function ${\sin }^{-1}x$) followed by algebraic manipulation and substitution to evaluate the resulting integral.

---

Solution

Let:

$I=\int x^2{\sin }^{-1}xdx$

Step 1: Apply Integration by Parts

Using the formula $\int udv=uv-\int vdu$, we choose:

• $u={\sin }^{-1}x$ (so that $du=\frac{1}{\sqrt{1-x^2}}dx$)
• $dv=x^{2}dx$ (so that $v=\frac{x^3}{3}$)

This gives:

$\begin{array}{rl}I& ={\sin }^{-1}x\cdot \frac{x^3}{3}-\int \frac{x^3}{3}\cdot \frac{1}{\sqrt{1-x^2}}dx\\ & =\frac{x^3}{3}{\sin }^{-1}x-\frac{1}{3}\int \frac{x^3}{\sqrt{1-x^2}}dx\end{array}$

Step 2: Simplify the Remaining Integral

We focus on $\int \frac{x^3}{\sqrt{1-x^2}}dx$. Rewrite $x^3$ as $x\cdot x^2$ and use the identity $x^2=1-(1-x^2)$:

$\begin{array}{rl}\int \frac{x^3}{\sqrt{1-x^2}}dx& =\int \frac{x\cdot x^2}{\sqrt{1-x^2}}dx\\ & =\int \frac{x(1-(1-x^2))}{\sqrt{1-x^2}}dx\\ & =\int \left(\frac{x}{\sqrt{1-x^2}}-\frac{x(1-x^2)}{\sqrt{1-x^2}}\right)dx\\ & =\int x(1-x^2{)}^{-1/2}dx-\int x(1-x^2{)}^{1/2}dx\end{array}$

Step 3: Evaluate Using Substitution

For both integrals, use the substitution $t=1-x^2$, which gives $dt=-2xdx$ or $xdx=-\frac{1}{2}dt$.

First integral: $\int x(1-x^2{)}^{-1/2}dx=-\frac{1}{2}\int t^{-1/2}dt=-\frac{1}{2}\cdot \frac{t^{1/2}}{1/2}=-t^{1/2}=-(1-x^2{)}^{1/2}$

Second integral: $\int x(1-x^2{)}^{1/2}dx=-\frac{1}{2}\int t^{1/2}dt=-\frac{1}{2}\cdot \frac{t^{3/2}}{3/2}=-\frac{1}{3}{t}^{3/2}=-\frac{1}{3}(1-x^2{)}^{3/2}$

Step 4: Combine Results

Substituting back into our expression for $I$:

$\begin{array}{rl}I& =\frac{x^3}{3}{\sin }^{-1}x-\frac{1}{3}\left[-(1-x^2{)}^{1/2}-\left(-\frac{1}{3}(1-x^2{)}^{3/2}\right)\right]+C\\ & =\frac{x^3}{3}{\sin }^{-1}x-\frac{1}{3}\left[-(1-x^2{)}^{1/2}+\frac{1}{3}(1-x^2{)}^{3/2}\right]+C\\ & =\frac{x^3}{3}{\sin }^{-1}x+\frac{1}{3}(1-x^2{)}^{1/2}-\frac{1}{9}(1-x^2{)}^{3/2}+C\end{array}$

Therefore, the final answer is:

$\boxed{\frac{x^3}{3}{\sin }^{-1}x+\frac{1}{3}{\left(1-x^2\right)}^{\frac{1}{2}}-\frac{1}{9}{\left(1-x^2\right)}^{\frac{3}{2}}+C}$

Or equivalently:

$\frac{x^3}{3}{\sin }^{-1}x+\frac{1}{3}\sqrt{1-x^2}-\frac{1}{9}(1-x^2{)}^{3/2}+C$

---

Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• Derivative of Inverse Sine: $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
• Algebraic Manipulation: Rewriting $x^3=x(1-(1-x^2))$ to separate integrable terms
• Substitution Method: $\int f(g(x))g^{\prime }(x)dx=\int f(u)du$ where $u=g(x)$
• Power Rule for Integration: $\int u^{n}du=\frac{u^{n+1}}{n+1}+C$ (for $n\ne -1$)

---

Summary of Steps

1. Set up integration by parts with $u={\sin }^{-1}x$ and $dv=x^{2}dx$ to reduce the problem to evaluating $\frac{1}{3}\int \frac{x^3}{\sqrt{1-x^2}}dx$
2. Rewrite the integrand by expressing $x^3=x(1-(1-x^2))$, which splits the fraction into $\frac{x}{\sqrt{1-x^2}}-x\sqrt{1-x^2}$
3. Separate into two integrals and use the substitution $t=1-x^2$ (noting that $xdx=-\frac{1}{2}dt$) for both
4. Apply the power rule to evaluate each integral: $\int t^{-1/2}dt$ and $\int t^{1/2}dt$
5. Back-substitute $t=1-x^2$ and combine all terms with the constant of integration $C$