Problem source

\begin{questionparts}
\item By a substitution of the form $y=k-x$ for suitable $k$, prove
that, for any function $\mathrm{f}$, 
\[
\int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\pi\int_{0}^{\frac{1}{2}\pi}\mathrm{f}(\sin x)\,\mathrm{d}x.
\]
Hence or otherwise evaluate 
\[
\int_{0}^{\pi}\frac{x}{2+\sin x}\,\mathrm{d}x.
\]
\item Evaluate 
\[
\int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t.
\]
{[}No credit will be given for numerical answers obtained by use of
a calculator.{]}  \end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
y = \pi - x, \d y = -\d x: && \int_0^{\pi} x f(\sin x) &= \int_{y = \pi}^{y = 0}(\pi - y) f(\sin(\pi-y))- \d y \\
&&&=  \int_0^{\pi} (\pi -y) f(\sin y) \d y \\
\Rightarrow && 2 \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi} f(\sin x) \d x \\
&&&= \pi \int_0^{\pi/2} f(\sin x ) \d x +   \pi \int_{\pi/2}^{\pi} f(\sin x ) \d x \\
&&&= \pi \int_0^{\pi/2} f(\sin x ) \d x +\pi \int_{y=\pi/2}^{y=0} f(\sin (\pi-y) ) (-\d y) \\
&&&= 2 \pi \int_0^{\pi/2} f(\sin x) \d x \\
\Rightarrow &&  \int_0^{\pi} x f(\sin x)\d  x &= \pi \int_0^{\pi/2} f(\sin x) \d x
\end{align*}

Therefore if $f(x) = \frac1{2+\sin x}$, letting $t = \tan \frac{x}{2}$ we have $\sin x = \frac{2 t}{1+t^2}, \frac{dt}{\d x} = \frac12 (1+t^2)$

\begin{align*}
&& \int_0^{\pi} \frac{x}{2 + \sin x } \d x &= \pi \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\
&&&= \pi \int_{t = 0}^{t = 1} \frac{1}{2+\frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\
&&&=\pi  \int_0^1 \frac{2}{2t^2+2t+2} \d t\\
&&&=\pi  \int_0^1 \frac{1}{(t+\tfrac12)^2 + \tfrac34} \d t\\
&&&= \pi \left [\frac{1}{\sqrt{3/4}} \tan^{-1} \frac{u}{\sqrt{3/4}} \right ]_{u=1/2}^{3/2} \\
&&&= \frac{2 \pi}{\sqrt{3}} \left ( \tan^{-1} \sqrt{3} - \tan^{-1} \frac1{\sqrt{3}} \right) \\
&&&= \frac{2 \pi}{\sqrt{3}} \left ( \frac{\pi}{3} - \frac{\pi}{6} \right) \\
&&&= \frac{\pi^2}{3\sqrt{3}}
\end{align*}

\item Let $u = (\sin^{-1} t)^2, \frac{\d u}{\d t} = 2(\sin^{-1} t) \frac{1}{\sqrt{1-t^2}}$


\begin{align*}
\int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t &= \int_{u=0}^{\pi^2/4} \frac12 \cos u \d u \\
&= \frac12 \sin \frac{\pi^2}{4}
\end{align*}

\end{questionparts}