If $$a\in \mathbb R,\int_{a-\pi}^{3\pi+a}|x-a-\pi|\sin(x/2)dx=-16$$ then a can be? I tried substituting $x-a=u$ and then breaking into two integrals removing modulus then used $\int \sin x=-\cos x,\int x\sin x=\sin x-x\cos x$

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    $\begingroup$ have you tried $u = x - a - \pi$ from $u = -2\pi$ to $u = 2\pi$? $\endgroup$ – GFauxPas Nov 30 '14 at 13:49
  • $\begingroup$ @GFauxPas nope, good observation. $\endgroup$ – RE60K Nov 30 '14 at 13:52

The evaluation of the integral on the left hand side of the integral equation

\begin{equation*} \int_{a-\pi }^{3\pi +a}\left\vert x-a-\pi \right\vert \sin \left( x/2\right)\tag{1} \,dx=-16 \end{equation*}

can be carried out starting with the substitution suggested by GFauxPas $y=x-a-\pi $ in a comment, [edit] splitting the integral into two, one for $-2\pi <y<0$ and the other for $0\leq y<2\pi $, and proceeding with the substitution $z=\frac{y+a}{2}$ and integration by parts [edit end]:

\begin{eqnarray*} -16 &=&\int_{a-\pi }^{3\pi +a}\left\vert x-a-\pi \right\vert \sin \left( x/2\right) \,dx \\ &=&\int_{-2\pi }^{2\pi }\left\vert y\right\vert \sin \left( \frac{y+a}{2}+ \frac{\pi }{2}\right) \,dy,\qquad\qquad y=x-a-\pi \\ &=&\int_{-2\pi }^{2\pi }\left\vert y\right\vert \cos \left( \frac{y+a}{2} \right) \,dy \\ &=&-\int_{-2\pi }^{0}y\cos \left( \frac{y+a}{2}\right) \,dy+\int_{0}^{2\pi }y\cos \left( \frac{y+a}{2}\right) \,dy, \\ &=&-\left[ 2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}\right] _{-2\pi }^{0} +\left[ 2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}\right] _{0}^{2\pi }\tag{$\ast$} \\ &=&-8\cos \frac{a}{2}+4\pi \sin \frac{a}{2}-8\cos \frac{a}{2}-4\pi \sin \frac{a}{2} \\ &=&-16\cos \frac{a}{2}, \end{eqnarray*}


\begin{eqnarray*} I(y) &=&\int y\cos \left( \frac{y+a}{2}\right) \,dy \\ &=&\int 2\left( 2z-a\right) \cos z\,dz,\qquad \qquad z=\frac{y+a}{2} \\ &=&4\int z\cos z\,dz-2a\int \cos z\,dz \end{eqnarray*}


\begin{eqnarray*} I(y) &=&4\left( z\sin z-\int \sin z\,dz\right) -2a\sin z \\ &=&\left( 4z-2a\right) \sin z+4\cos z \\ &=&2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}.\tag{$\ast$} \end{eqnarray*}

Hence $(1)$ is equivalent to the simple trigonometric equation

\begin{equation*} \cos \frac{a}{2}=1,\tag{2} \end{equation*}

whose solution is

\begin{equation*} a=4k\pi ,\text{ }k\in\mathbb{Z}.\tag{3} \end{equation*}


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