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

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 $x$:
$$
I=\frac{1}{8}\arcsin x+\frac{1}{8}x(2x^2-1)\sqrt{1-x^2}+c
$$
