Problem source

Evaluate 
\begin{questionparts}
\item  ${\displaystyle \int_{-\pi}^{\pi}\left|\sin x\right|\,\mathrm{d}x,}$
\item ${\displaystyle \int_{-\pi}^{\pi}\sin\left|x\right|\,\mathrm{d}x},$ 
\item ${\displaystyle \int_{-\pi}^{\pi}x\sin x\,\mathrm{d}x},$ 
\item ${\displaystyle \int_{-\pi}^{\pi}x^{10}\sin x\,\mathrm{d}x.}$
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
\int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\
&= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\
&= 1-(-1)+(1)-(-1) \\
&= 4
\end{align*}

\item \begin{align*}
\int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\
&= 4
\end{align*}
\item \begin{align*}
\int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\
&= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\
&= 2\pi
\end{align*}
\item \begin{align*}
\int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0
\end{align*}
\end{questionparts}