Problem source

Let 
\[
u_{n}=\int_{0}^{\frac{1}{2}\pi}\sin^{n}t\,\mathrm{d}t
\]
for each integer $n\geqslant0$. By integrating 
\[
\int_{0}^{\frac{1}{2}\pi}\sin t\sin^{n-1}t\,\mathrm{d}t
\]
 by parts, or otherwise, obtain a formula connecting $u_{n}$ and
$u_{n-2}$ when $n\geqslant2$ and deduce that 
\[
nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}
\]
for all $n\geqslant2$. Deduce that 
\[
nu_{n}u_{n-1}=\tfrac{1}{2}\pi.
\]
Sketch graphs of $\sin^{n}t$ and $\sin^{n-1}t$, for $0\leqslant t\leqslant\frac{1}{2}\pi,$
on the same diagram and explain why $0 < u_{n} < u_{n-1}.$ By using the 
result of the previous paragraph show that 
\[
nu_{n}^{2} < \tfrac{1}{2}\pi < nu_{n-1}^{2}
\]
for all $n\geqslant1$. Hence show that 
\[
\left(\frac{n}{n+1}\right)\tfrac{1}{2}\pi < nu_{n}^{2} < \tfrac{1}{2}\pi
\]
and deduce that $nu_{n}^{2}\rightarrow\tfrac{1}{2}\pi$ as $n\rightarrow\infty$.

Solution source

\begin{align*}
&& u_n &= \int_0^{\tfrac12 \pi} \sin^{n} t \, \d t \\
&& &=  \int_0^{\tfrac12 \pi} \sin t \sin^{n-1} t \, \d t \\
&& &= \left [ -\cos t \sin^{n-1} t  \right]_0^{\tfrac12 \pi} + \int_0^{\tfrac12 \pi} \cos t (n-1) \sin^{n-2} t \cos t \d t \\
&& &= 0 + (n-1)\int_0^{\tfrac12 \pi} \cos^2 t \sin^{n-2} t \d t \\
&& &= (n-1) \int_0^{\tfrac12 \pi}(1-\sin^2 t) \sin^{n-2} t \d t \\
&& &= (n-1)u_{n-2} - (n-1)u_n \\
\Rightarrow && n u_n &= (n-1)u_{n-2} \\
\end{align*}

Mutplying both sides by $u_{n-1}$ we obtain $nu_{n}u_{n-1}=\left(n-1\right)u_{n-1}u_{n-2}$.

Therefore $nu_nu_{n-1}$ is constant, ie is equal to $\displaystyle u_1u_0 = \int_0^{\tfrac12 \pi} \sin^{1} t \, \d t \int_0^{\tfrac12 \pi} \sin^{0} t \, \d t  = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(#1)^5};
    \def\functiong(#1){sin(#1)^6};
    \def\xl{-5};
    \def\xu{100};
    \def\yl{-0.2};
    \def\yu{1.25};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:90, samples=100] 
            plot (\x, {\functionf(\x)});
        \draw[thick, red, smooth, domain=0:90, samples=100] 
            plot (\x, {\functiong(\x)});
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Since $0 < \sin t < 1$ for $t \in (0, \tfrac{\pi}{2})$ we must have $0 < \sin^n t <  \sin^{n-1} t$, in particular $0 < u_n < u_{n-1}$

Therefore

\begin{align*}
&& nu_{n}u_{n-1} &= \tfrac{1}{2}\pi \\
\Rightarrow && nu_n u_n &< \tfrac{1}{2}\pi  \tag{$u_n < u_{n-1}$} \\
\Rightarrow && nu_{n-1} u_{n-1} &> \tfrac{1}{2}\pi  \tag{$u_n < u_{n-1}$} \\
\Rightarrow && nu_n^2 &< \tfrac12 \pi < n u_{n-1}^2
\end{align*}

However we also have $\tfrac12 \pi < (n+1)u_n^2$ (by considering the next inequality), so 

$\left ( \frac{n}{n+1}\right) \tfrac12 \pi < n u_n^2 < \tfrac12 \pi$

but since as $n \to \infty$ the right hand bound is constant and the left hand bound tends to $\tfrac12 \pi$ therefore $n u_n^2 \to \tfrac12 \pi$