|
\[
I_4 = \displaystyle\int_{0}^{\pi}
\max\left(\left|2\sin(x)\right|,\left|2\cos(2x)-1\right| \right)^2\cdot
\min\left(\left|\sin(2x)\right|,\left|\cos(3x)\right| \right)^2 dx
\]
A primera vista ¨¦s una integral molt complicada. No exactament dificil, per¨° amb molta feina associada.
Primer caldr¨¤ calcular els punts de tall de cada parella de funcions per determinar en quins intervals la m¨¤xima (o la m¨ªnima) s'ha de calcular.
\[
\left|2\sin(x)\right|=\left|2\cos(2x)-1\right|
\hspace{20pt}
\left|\sin(2x)\right|=\left|\cos(3x)\right|
\]
podem eliminar els valors absoluts considerant els casos possible i fent
\[
\pm2\sin(x)=2\cos(2x)-1
\hspace{20pt}
\pm\sin(2x)=\cos(3x)
\]
Ara, utilitzant f¨®mules conegudes
\[
\begin{array}{cc}
\pm2\sin(x)=2(1-2\sin^2x)-1
&
\pm2\sin(x)\cos(x)=\cos(x)\left(1-4\sin^2x\right)\\
\pm2\sin(x)=1-4\sin^2x)
&
\pm2\sin(x)\cos(x)-\cos(x)\left(1-4\sin^2x\right)=0\\
4\sin^2x\pm2\sin(x)-1=0
&
\cos(x)\left(\pm2\sin(x)-\left(1-4\sin^2x\right)\right)=0\\
&
\cos(x)\left(4\sin^2x\pm2\sin(x)-1\right)=0
\end{array}
\]
Quasi tenen els mateixos punts de tall que simplifica un xic el problema
\[
4\sin^2x\pm2\sin(x)-1=0
\]
\[
\sin x=\dfrac{-(\pm2)\pm\sqrt{(\pm2)^2-4\cdot4\cdot(-1)}}{2\cdot4}=\dfrac{\mp2\pm\sqrt{4+16}}{8}=\dfrac{\mp2\pm\sqrt{20}}{8}=\dfrac{\mp1\pm\sqrt{5}}{4}
\]
Com que estem considerant el interval \([0,\pi]\) i tenin en compte les multiplicitats
\[
x=\sin^{-1}\left(\dfrac{-1+\sqrt{5}}{4}\right)=\dfrac{\pi}{10}=18^\circ \hspace{16pt}\implies\hspace{16pt} x=\pi-\dfrac{\pi}{10}=\dfrac{9\pi}{10}=162^\circ
\]
\[
x=\sin^{-1}\left(\dfrac{+1+\sqrt{5}}{4}\right)=\dfrac{3\pi}{10}=54^\circ \hspace{16pt}\implies\hspace{16pt} x=\pi-\dfrac{3\pi}{10}=\dfrac{7\pi}{10}=126^\circ
\]
Per les funcions del m¨ªnim tamb¨¦ hi ha un altre punt de intersecci¨®:
\[
\cos(x)=0 \hspace{16pt}\implies\hspace{16pt} x=\dfrac{\pi}{2}
\]
per¨° com ¨¦s pot veure al gr¨¤fic els dos intervals adjacents es poden ajuntar per que les funcionstenen el mateix comportament
\[
\left[\dfrac{3\pi}{10},\dfrac{\pi}{2}\right)\cup
\left[\dfrac{\pi}{2},\dfrac{7\pi}{10}\right)=
\left[\dfrac{3\pi}{10},\dfrac{7\pi}{10}\right)
\]
tenim cinc intervals de integraci¨®:
\[
\left[0,\dfrac{\pi}{10}\right)
\hspace{16pt}
\left[\dfrac{\pi}{10},\dfrac{3\pi}{10}\right)
\hspace{16pt}
\left[\dfrac{3\pi}{10},\dfrac{7\pi}{10}\right)
\hspace{16pt}
\left[\dfrac{7\pi}{10},\dfrac{9\pi}{10}\right)
\hspace{16pt}
\left[\dfrac{9\pi}{10},\pi\right]
\]
Per tant, mirant grafics aproximats o calculant el valor de cada funci¨® en cada integrand
\[
\begin{array}{rl}
\displaystyle\int_{0}^{\pi}
\max\left(\left|2\sin(x)\right|,\left|2\cos(2x)-1\right| \right)^2\cdot
\min\left(\left|\sin(2x)\right|,\left|\cos(3x)\right| \right)^2dx =
& \displaystyle\int_{0}^{\pi/10} \left(2\cos(2x)-1\right)^2\left(\sin(2x)\right)^2dx\\
+& \displaystyle\int_{\pi/10}^{3\pi/10} \left(2\sin(x)\right)^2\left(\cos(3x)\right)^2dx\\
+& \displaystyle\int_{3\pi/10}^{7\pi/10} \left(2\cos(2x)-1\right)^2\left(\sin(2x)\right)^2dx\\
+& \displaystyle\int_{7\pi/10}^{9\pi/10} \left(2\sin(x)\right)^2\left(\cos(3x)\right)^2dx\\
+& \displaystyle\int_{9\pi/10}^{\pi} \left(2\cos(2x)-1\right)^2\left(\sin(2x)\right)^2dx
\end{array}
\]
Sembla que hi ha dos integrands diferents, per¨° son iguals
\[
\begin{array}{rl}
\left(2\cos(2x)-1\right)^2\left(\sin(2x)\right)^2
&=\left(2\left[2\cos^2(x)-1\right]-1\right)^2\left(2\sin(x)\cos(x)\right)^2\\
&=4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2
\end{array}
\]
\[
\begin{array}{rl}
\left(2\sin(x)\right)^2\left\{\cos(3x)\right\}^2
&=4\sin^2(x)\left\{\cos(x)\left[4\cos^2(x)-3\right]\right\}^2\\
&=4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2
\end{array}
\]
Per tant els intervals es poden ajuntar
\[
\begin{array}{rl}
I_4= & \displaystyle\int_{0}^{\pi/10} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx\\
+& \displaystyle\int_{\pi/10}^{3\pi/10} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx\\
+& \displaystyle\int_{3\pi/10}^{7\pi/10} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx\\
+& \displaystyle\int_{7\pi/10}^{9\pi/10} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx\\
+& \displaystyle\int_{9\pi/10}^{\pi} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx
= \displaystyle\int_{0}^{\pi} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx
\end{array}
\]
Aix¨ª la integral s'ha transformat en
\[
I_4 = \displaystyle\int_{0}^{\pi} 4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2dx
\]
Treballant el integrand
\[
\begin{array}{rl}
4\sin^2(x)\cos^2(x)\left[4\cos^2(x)-3\right]^2
&= \sin^2(2x)\left[4\dfrac{1+\cos(2x)}{2}-3\right]^2\\
&= \dfrac{1-\cos(4x)}{2}\left[4\dfrac{1+\cos(2x)}{2}-3\right]^2\\
&= \dfrac{1}{2}\left(1-\cos(4x)\right)\left[2\cos(2x)-1\right]^2\\
&= \dfrac{1}{2}\left(1-\cos(4x)\right)\left[4\cos^2(2x)-4\cos(2x)+1\right]\\
&= \dfrac{1}{2}\left(1-\cos(4x)\right)\left[4\dfrac{1+\cos(4x)}{2}-4\cos(2x)+1\right]\\
&= \dfrac{1}{2}\left(1-\cos(4x)\right)\left[2+2\cos(4x)-4\cos(2x)+1\right]\\
&= \dfrac{1}{2}\left(1-\cos(4x)\right)\left[3+2\cos(4x)-4\cos(2x)\right]\\
&= \dfrac{1}{2}\left(3+2\cos(4x)-4\cos(2x) -3\cos(4x)-2\cos^2(4x)+4\cos(2x)\cos(4x)\right)\\
&= \dfrac{1}{2}\left[3+2\cos(4x)-4\cos(2x) -3\cos(4x)-(1+\cos(8x))+4\dfrac{1}{2}\left(\cos(6x)+\cos(2x)\right)\right]\\
&= \dfrac{1}{2}\left[3+2\cos(4x)-4\cos(2x) -3\cos(4x)-1-\cos(8x)+2\cos(6x)+2\cos(2x)\right]\\
&= \dfrac{1}{2}\left[2-2\cos(2x) -\cos(4x)+2\cos(6x) -\cos(8x)\right]\\
&= 1-\cos(2x) -\dfrac{1}{2}\cos(4x)+\cos(6x) -\dfrac{1}{2}\cos(8x)
\end{array}
\]
llavors, finalment
\[
\begin{array}{rl}
I_4&=\displaystyle\int_{0}^{\pi}\left(1-\cos(2x) -\dfrac{1}{2}\cos(4x)+\cos(6x) -\dfrac{1}{2}\cos(8x)\right)dx\\
&=\left.x-\dfrac{1}{2}\sin(2x)-\dfrac{1}{8}\sin(4x)+\dfrac{1}{6}\sin(6x)-\dfrac{1}{16}\sin(8x)\right|_{0}^{\pi}\\
&=\pi-\dfrac{1}{2}\sin(2\pi)-\dfrac{1}{8}\sin(4\pi)+\dfrac{1}{6}\sin(6\pi)-\dfrac{1}{16}\sin(8\pi)- (zeros)\\
&=\pi
\end{array}
\]
perque tots aquests sinus son zero i els sinus de zero tamb¨¦ ho son.
\[
I_4 = \displaystyle\int_{0}^{\pi}
\max\left(\left|2\sin(x)\right|,\left|2\cos(2x)-1\right| \right)^2\cdot
\min\left(\left|\sin(2x)\right|,\left|\cos(3x)\right| \right)^2 dx = \pi
\]
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