Integral exercise mixed with trig and min , max asked in Imad Zak math group by Dan Sitaru

Exercise:

Solve in$\left( 0,\pi  \right)$ such that, $\int_{0}^{\pi }{\sin \left( \max \left( x,t \right) \right)\cos \left( \min \left( x,t \right) \right)dt}=0$

Solution: $\max \left( x,t \right)=\frac{x+t+\left| x-t \right|}{2}$ and $\min \left( x,t \right)=\frac{x+t-\left| x-t \right|}{2}$

So $\int_{0}^{\pi }{\sin \left( \max \left( x,t \right) \right)\cos \left( \min \left( x,t \right) \right)dt}=\int_{0}^{\pi }{\sin \left( \frac{x+t+\left| x-t \right|}{2} \right)\cos \left( \frac{x+t-\left| x-t \right|}{2} \right)dt}=0$

 We know that $\sin \left( x \right)\cos \left( y \right)=\frac{1}{2}\left( \sin \left( x+y \right)+\sin \left( x-y \right) \right)$

So $\sin \left( \frac{x+t+\left| x-t \right|}{2} \right)\cos \left( \frac{x+t-\left| x-t \right|}{2} \right)=\frac{1}{2}\sin \left( \frac{2\left( x+t \right)}{2} \right)+\frac{1}{2}\sin \left( \frac{2\left| x-t \right|}{2} \right)$

$\Leftrightarrow \sin \left( \frac{x+t+\left| x+t \right|}{2} \right)\cos \left( \frac{x+t-\left| x-t \right|}{2} \right)=\frac{1}{2}\sin \left( x+t \right)+\frac{1}{2}\sin \left| x-t \right|$

$\Leftrightarrow \int_{0}^{\pi }{\sin \left( \frac{x+t+\left| x-t \right|}{2} \right)\cos \left( \frac{x+t-\left| x-t \right|}{2} \right)dt}=\frac{1}{2}\int_{0}^{\pi }{\sin \left( x+t \right)dt+\frac{1}{2}\int_{0}^{\pi }{\sin \left| x-t \right|dt}=0}$

Observe that ,

    $\sin \left| x-t \right|=\left\{ \begin{align}
  & \sin \left( x-t \right)\,\,,\,\,x>t \\
 & -\sin \left( x-t \right)\,\,,x<t \\
\end{align} \right.$

So $\frac{1}{2}\left( \int_{0}^{\pi }{\sin \left( x+t \right)dt+\int_{0}^{x}{\sin \left( x-t \right)dt-\int_{x}^{\pi }{\sin \left( x-t \right)dt}}} \right)=0$

Let $u=x+t\Leftrightarrow du=dt\,\,\And \,\,v=x-t\Leftrightarrow dv=-dt$

$\Leftrightarrow \int_{x}^{x+\pi }{\sin udu}-\int_{x}^{0}{\sin vdv+\int_{0}^{x-\pi }{\sin v\,dv}=0}$


$\Leftrightarrow \left( \cos u \right)_{x}^{x+\pi }-\left( \cos v \right)_{x}^{0}+\left( \cos v \right)_{0}^{x-\pi }=0$

$\Leftrightarrow \cos \left( \pi +x \right)-\cos x-\cos 0+\cos x+\cos \left( x-\pi  \right)-\cos 0=0$

$\Leftrightarrow -\cos x-\cos x-1+\cos x-\cos x-1=0$

$\Leftrightarrow -2\cos x-2=0\Leftrightarrow -2\left( \cos x+1 \right)=0\Leftrightarrow \cos x+1=0\Leftrightarrow \cos x=-1$

So $x=\pi \notin \left( 0,\pi  \right)$ thus this equation has no solution in the interval $\left( 0,\pi  \right)$

*_____________________
The idea of $max(x,y) $ & $min(x,t)$ credit to Diego Alvariz