Problem source

If $\mathrm{Q}$ is a polynomial, $m$ is an integer, $m\geqslant1$
and $\mathrm{P}(x)=(x-a)^{m}\mathrm{Q}(x),$ show that \[
\mathrm{P}'(x)=(x-a)^{m-1}\mathrm{R}(x)
\]
where $\mathrm{R}$ is a polynomial. Explain why $\mathrm{P}^{(r)}(a)=0$
whenever $1\leqslant r\leqslant m-1$. ($\mathrm{P}^{(r)}$ is the
$r$th derivative of $\mathrm{P}.$)
If 
\[
\mathrm{P}_{n}(x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n}
\]
for $n\geqslant1$ show that $\mathrm{P}_{n}$ is a polynomial of
degree $n$. By repeated integration by parts, or otherwise, show
that, if $n-1\geqslant m\geqslant0,$ 
\[
\int_{-1}^{1}x^{m}\mathrm{P}_{n}(x)\,\mathrm{d}x=0
\]
and find the value of 
\[
\int_{-1}^{1}x^{n}\mathrm{P}_{n}(x)\,\mathrm{d}x.
\]
{[}\textbf{Hint. }\textit{You may use the formula 
\[
 \int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t\,\mathrm{d}t=\frac{(2^{2n})(n!)^{2}}{(2n+1)!}
\]
 without proof if you need it. However some ways of doing this question
do not use this formula.}{]}

Solution source

\begin{align*}
&& P(x) &= (x-a)^mQ(x) \\
\Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\
&&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\
&&&= (x-a)^{m-1}R(x)
\end{align*}

Therefore $P^{(r)}(a) = 0$ for $1 \leq r \leq m-1$ since each time we differentiate we will have a factor of $(x-a)^{m-r}$ which is zero when we evaluate at $x = a$.

If $P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n$ then we are differentiating a degree $2n$ polynomial $n$ times. Each time we differentiate we reduce the degree by $1$, therefore the degree of $P_n$ is $n$.

\begin{align*}
&& \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\
&&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\
&&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\
&&&=  m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\
&&& \cdots \\
&&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\
&&&= 0
\end{align*}

If $n = m$, we have

\begin{align*}
&& \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\
&&  &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\
x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\
&&&=  2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\
&&&=  2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\
&&&=  \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\
\end{align*}