Evaluating an Integral with a $\sqrt{1-x^2}$ in it

Question

Below is a problem I did but my answer does not match the book's table of integrals. I would like to know where I went wrong.
Thanks
Bob
Problem:
Evaluate the following integral: \begin{eqnarray*} \int x^2 \sqrt{ 1 - x^2 } \,\, dx \\ \end{eqnarray*}
Answer
\begin{eqnarray*} \text{Let } I &=& \int x^2 \sqrt{ 1 - x^2 } \,\, dx \\ \\ \text{Let }\sin u &=& x \\ dx &=& \cos{u} \, du \\ I &=& \int \sin^2{u} \sqrt{1 - \sin^2u} \,\, \cos{u} \, du \\ I &=& \int \sin^2{u} \cos^2{u} \,\, du \\ \end{eqnarray*} Now recall the following two standard identities: \begin{eqnarray*} \sin^2{\theta} &=& \frac{1 - \cos{2 \theta}}{2} \\ \cos^2{\theta} &=& \frac{1 + \cos{2 \theta}}{2} \\ \end{eqnarray*} Now applying the above two identities we have: \begin{eqnarray*} I &=& \int \frac{(1 - \cos{2 u})(1 + \cos{2 u})}{4} \,\, du \\ 4I &=& \int 1 - \cos^2{2u} \,\, du = \int \sin^2{2u} \,\, du \\ 4I &=& \int \frac{1 - \cos{4u}}{2} \,\, d\theta \\ 8I &=& \int 1 - \cos{4u} \,\, du \\ \int \cos{4u} \,\, du &=& \frac{\sin{4 u}}{4} + C_1 \\ \end{eqnarray*} Now recall the following two standard identities: \begin{eqnarray*} \sin{2 \theta} &=& 2 \sin{\theta} \cos{\theta} \\ \cos{2 \theta} &=& 1 - 2 \sin^2{\theta} \\ \end{eqnarray*} \begin{eqnarray*} \sin{4 u} &=& 2 \sin{2 u} \cos{2 u} = 4 \sin u \cos u ( 1 - 2 \sin^2 u) \\ \sin{4 u} &=& 4 \sin u \cos u - 8 \sin^3 u \cos u \\ \sin{4 u} &=& (4 \sin u - 8 \sin^3 u ) ( 1 - \sin^2 u)^\frac{1}{2} \\ \int \cos{4u} \,\, du &=& ( \sin u - 2 \sin^3 u ) ( 1 - \sin^2 u)^\frac{1}{2} + C_2 \\ 8I &=& \sin^{-1} x - (x - 2x^3)\sqrt{1 - x^2} + C_2 \\ I &=& \frac{1}{8}\sin^{-1} x - \frac{(x - 2x^3)\sqrt{1 - x^2}}{8} + C \\ \end{eqnarray*} However, the book's answer is: \begin{eqnarray*} \int x^2 \sqrt{1 - x^2} \,\, dx &=& \frac{1}{8} \sin^{-1}{x} - \frac{x\sqrt{1-x^2}(1-2x^2)}{8} + C \\ \end{eqnarray*}

Answer 1

Let's try a different method: $$ x^2\sqrt{1-x^2}=\frac{x^2-x^4}{\sqrt{1-x^2}}=(x^3-x)\frac{-x}{\sqrt{1-x^2}} $$ Let's do integration by parts: \begin{align} I&=\int x^2\sqrt{1-x^2}\,dx\\[6px] &=\int(x^3-x)\frac{-x}{\sqrt{1-x^2}}\,dx\\[6px] &=(x^3-x)\sqrt{1-x^2}-\int(3x^2-1)\sqrt{1-x^2}\,dx\\[6px] &=(x^3-x)\sqrt{1-x^2}-3I+\int\sqrt{1-x^2}\,dx \end{align} The last integral is well known: $$ \int\sqrt{1-x^2}\,dx=\frac{1}{2}\arcsin x+\frac{1}{2}x\sqrt{1-x^2} $$ so we obtain $$ 4I=\frac{1}{2}\arcsin x+\frac{1}{2}(2x^3-x)\sqrt{1-x^2}+c $$ which is the same as your solution, but also the same as the book's, because we can collect $x$: $$ I=\frac{1}{8}\arcsin x+\frac{1}{8}x(2x^2-1)\sqrt{1-x^2}+c $$