Exercise:
Evaluate, $\int_{0}^{\pi }{\frac{x\sin x}{\cos 4x+4\cos 2x+11}dx}$
Solution: we know that $\cos 2\theta =2{{\cos }^{2}}\theta -1$ then
$\cos 4x=\cos \left( 2\left( 2x \right) \right)=2{{\cos }^{2}}2x-1$ and $\cos 2x=2{{\cos }^{2}}x-1$
Hence $\cos 4x+4\cos 2x+11=2{{\left( 2{{\cos }^{2}}x-1 \right)}^{2}}-1+4\left( 2{{\cos }^{2}}x-1 \right)+11$
$=2\left( 4{{\cos }^{4}}x+1-4{{\cos }^{2}}x \right)-1+8{{\cos }^{2}}x-4+11$
$=8{{\cos }^{4}}x-8{{\cos }^{2}}x+2-1+8{{\cos }^{2}}x-4+11=8{{\cos }^{4}}x+8=8\left( {{\cos }^{4}}x+1 \right)$
Hence $\int_{0}^{\pi }{\frac{x\sin x}{\cos 4x+4\cos 2x+11}dx}=\frac{1}{8}\int_{0}^{\pi }{\frac{x\sin x}{{{\cos }^{4}}x+1}dx}$
But $\int_{0}^{\pi }{\frac{x\sin x}{{{\cos }^{4}}x+1}dx}=\int_{0}^{\pi }{\frac{\left( \pi -x \right)\sin \left( \pi -x \right)}{{{\cos }^{4}}\left( \pi -x \right)+1}dx}=\int_{0}^{\pi }{\frac{\left( \pi -x \right)\sin x}{{{\cos }^{4}}x+1}dx}$
$=\int_{0}^{\pi }{\frac{\pi \sin x}{{{\cos }^{4}}x+1}dx}-\int_{0}^{\pi }{\frac{x\sin x}{{{\cos }^{4}}x+1}dx}$
Thus $\int_{0}^{\pi }{\frac{x\sin x}{{{\cos }^{4}}x+1}dx}=\frac{\pi }{2}\int_{0}^{\pi }{\frac{\sin x}{{{\cos }^{4}}x+1}dx}$
Therefore, $\int_{0}^{\pi }{\frac{x\sin x}{\cos 4x+4\cos 2x+11}dx}=\frac{\pi }{16}\int_{0}^{\pi }{\frac{\sin x}{{{\cos }^{4}}x+1}dx}$
Let $u=\cos x\Leftrightarrow du=-\sin xdx$ and $u\left( 0 \right)=1\,\,,\,\,u\left( \pi \right)=-1$ , then
$\int_{0}^{\pi }{\frac{\sin x}{{{\cos }^{4}}x+1}dx}=-\int_{1}^{-1}{\frac{du}{{{u}^{4}}+1}=\int_{-1}^{1}{\frac{du}{{{u}^{4}}+1}}}$
We know that , $\frac{1}{{{u}^{4}}+1}=\frac{2}{2\left( {{u}^{4}}+1 \right)}=\frac{1+{{u}^{2}}+1-{{u}^{2}}}{2\left( {{u}^{4}}+1 \right)}=\frac{1}{2}\left[ \frac{{{u}^{2}}+1}{{{u}^{4}}+1}-\frac{{{u}^{2}}-1}{{{u}^{4}}+1} \right]$
$=\frac{1}{2}\left[ \frac{\frac{{{u}^{2}}+1}{{{u}^{2}}}}{\frac{{{u}^{4}}+1}{{{u}^{2}}}}-\frac{\frac{{{u}^{2}}-1}{{{u}^{2}}}}{\frac{{{u}^{4}}+1}{{{u}^{2}}}} \right]=\frac{1}{2}\left[ \frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}-\frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}} \right]$
Hence $\int_{-1}^{1}{\frac{1}{{{u}^{4}}+1}du}=\frac{1}{2}\int_{-1}^{1}{\frac{1+\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}du}-\frac{1}{2}\int_{-1}^{1}{\frac{1-\frac{1}{{{u}^{2}}}}{{{u}^{2}}+\frac{1}{{{u}^{2}}}}du}$
Let $w=u+\frac{1}{u}\,\,\And \,\,v=u-\frac{1}{u}\Leftrightarrow dw=1-\frac{1}{{{u}^{2}}}\,du\,\,\And \,\,dv=1+\frac{1}{{{u}^{2}}}du$ also
${{w}^{2}}={{u}^{2}}+\frac{1}{{{u}^{2}}}+2\Leftrightarrow {{w}^{2}}-2={{u}^{2}}+\frac{1}{{{u}^{2}}}$ and ${{v}^{2}}={{u}^{2}}+\frac{1}{{{u}^{2}}}-2\Leftrightarrow {{v}^{2}}+2={{u}^{2}}+\frac{1}{{{u}^{2}}}$
$\int{\frac{du}{{{u}^{4}}+1}}=\frac{1}{2}\int{\frac{dv}{{{v}^{2}}+2}-\frac{1}{2}\int{\frac{dw}{{{w}^{2}}-2}}}$ but ${{w}^{2}}-2=\left( w-\sqrt{2} \right)\left( w+\sqrt{2} \right)$
Hence $\frac{1}{{{w}^{2}}-2}=\frac{A}{w-\sqrt{2}}+\frac{B}{w+\sqrt{2}}=\frac{\left( A+B \right)w+\left( A-B \right)\sqrt{2}}{{{w}^{2}}-2}$
So $A+B=0\,\,,\,\,A-B=\frac{1}{\sqrt{2}}\Leftrightarrow A=\frac{1}{2\sqrt{2}}\,\,\And B=-\frac{1}{2\sqrt{2}}$
Hence \[\int{\frac{dw}{{{w}^{2}}-2}=\frac{1}{2\sqrt{2}}\left[ \int{\frac{dw}{w-\sqrt{2}}-\int{\frac{dw}{w+\sqrt{2}}}} \right]}\]
$=\frac{1}{2\sqrt{2}}\ln \left| \frac{w-\sqrt{2}}{w+\sqrt{2}} \right|+c=\frac{1}{2\sqrt{2}}\ln \left| \frac{{{u}^{2}}-\sqrt{2}u+1}{{{u}^{2}}+\sqrt{2}u+1} \right|+c$
Thus \[\int_{-1}^{1}{\frac{1-1/{{u}^{2}}}{{{u}^{2}}+1/{{u}^{2}}}du=\frac{1}{2\sqrt{2}}}\left[ \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \right)-\ln \left( \frac{2+\sqrt{2}}{2-\sqrt{2}} \right) \right]=\frac{1}{2\sqrt{2}}\ln \left( 17-12\sqrt{2} \right)\] (*)
Also $\int{\frac{dv}{{{v}^{2}}+2}}=\int{\frac{dv}{{{v}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}}=\frac{1}{\sqrt{2}}\int{\frac{dy}{{{y}^{2}}+1}}\,\,\,,\,\,\,v=\sqrt{2}y\Leftrightarrow dv=\sqrt{2}dy$
$=\frac{1}{\sqrt{2}}\arctan \left( y \right)+c=\frac{1}{\sqrt{2}}\arctan \left( \frac{v}{\sqrt{2}} \right)+c$
But $\frac{1+1/{{u}^{2}}}{{{u}^{2}}+1/{{u}^{2}}}$ is discontinuous at $u=0$ ,then
Hence $\int_{-1}^{1}{\frac{1+1/{{u}^{2}}}{{{u}^{2}}+1/{{u}^{2}}}du}=\int_{-1}^{0}{\frac{1+1/{{u}^{2}}}{{{u}^{2}}+1/{{u}^{2}}}du}+\int_{0}^{1}{\frac{1+1/{{u}^{2}}}{{{u}^{2}}+1/{{u}^{2}}}du}$
$=\frac{1}{\sqrt{2}}\arctan \left( \frac{u-1/u}{\sqrt{2}} \right)_{-1}^{0}+\frac{1}{\sqrt{2}}\arctan \left( \frac{u-1/u}{\sqrt{2}} \right)_{0}^{1}$
$=\frac{\pi }{2\sqrt{2}}+\frac{\pi }{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}\,\,\,\,,\,\,\,\underset{u\to 0}{\mathop{\lim }}\,\arctan \left( \frac{u-1/u}{\sqrt{2}} \right)=\arctan \left( \infty \right)=\frac{\pi }{2}$ (**)
Therefore, $\int_{-1}^{1}{\frac{du}{{{u}^{4}}+1}=\frac{\pi }{2\sqrt{2}}-\frac{1}{2}\left( \frac{1}{2\sqrt{2}}\ln \left( 17-12\sqrt{2} \right) \right)=\frac{2\pi -\ln \left( 17-12\sqrt{2} \right)}{4\sqrt{2}}}$
Thus $\int_{0}^{\pi }{\frac{x\sin x}{\cos 4x+4\cos 2x+11}dx}=\frac{\pi \left( 2\pi -\ln \left( 17-12\sqrt{2} \right) \right)}{64\sqrt{2}}$ Q.E.D
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