Question #312407

If In=int_0 to ♾️ {(e^-x)(sin^n) (x) dx } , prove that (1+n^2)In=n(n-1)In-2 for n≥2 .

1
Expert's answer
2022-03-18T09:25:57-0400

Solution

In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \


We solve the given integral by integration by parts method


u = {e^{ - x}}\ and dv = \sin \left( {nx} \right)\,dx\


du = - {e^{ - x}}dx\ and v=cos(nx)nv = - \frac{{\cos \left( {nx} \right)}}{n}


Then 


In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \


In=[excos(nx)n]00(cos(nx)n)(ex)dxIn = \left[ { - \frac{{{e^{ - x}}\cos \left( {nx} \right)}}{n}} \right]_0^\infty - \int\limits_0^\infty {\left( { - \frac{{\cos \left( {nx} \right)}}{n}} \right)\left( { - {e^{ - x}}} \right)\,dx}


In=1n[cos(nx)ex]01n0excos(nx)dxIn = - \frac{1}{n}\left[ {\frac{{\cos \left( {nx} \right)}}{{{e^x}}}} \right]_0^\infty - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}


In=1n[cos(n())ecos(n(0))e(0)]1n0excos(nx)dxIn = - \frac{1}{n}\left[ {\frac{{\cos \left( {n\left( \infty \right)} \right)}}{{{e^\infty }}} - \frac{{\cos \left( {n\left( 0 \right)} \right)}}{{{e^{\left( 0 \right)}}}}} \right] - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}


In=1n[01]1n0excos(nx)dxIn = - \frac{1}{n}\left[ {0 - 1} \right] - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}


In=1n1n0excos(nx)dxIn = \frac{1}{n} - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}


 \begin{array}{l} In = \frac{1}{n} - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \\ \end{array}\


 \begin{array}{l} In = \frac{1}{n} - \frac{1}{n}\left( I \right) & & & ... & \left( 1 \right) \end{array}\


Now consider


I = \int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \


Solving by integration by parts 


u=exu = {e^{ - x}} and  dv=cos(nx)dxdv = \cos \left( {nx} \right)\,dx


du = - {e^{ - x}}dx\ and  v=sin(nx)nv = \frac{{\sin \left( {nx} \right)}}{n}


Then


I=[exsin(nx)n]00exsin(nx)ndxI = \left[ {\frac{{{e^{ - x}}\sin \left( {nx} \right)}}{n}} \right]_0^\infty - \int\limits_0^\infty {\frac{{ - {e^{ - x}}\sin \left( {nx} \right)}}{n}dx}


I=1n[sin(nx)ex]0+1n0exsin(nx)dxI = \frac{1}{n}\left[ {\frac{{\sin \left( {nx} \right)}}{{{e^x}}}} \right]_0^\infty + \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)dx}


I=1n[00]0+1n(In)I = \frac{1}{n}\left[ {0 - 0} \right]_0^\infty + \frac{1}{n}\left( {In} \right)


I=1n(In)I = \frac{1}{n}\left( {In} \right)



Hence from (1), we have 


In=1n1n(1n(In))In = \frac{1}{n} - \frac{1}{n}\left( {\frac{1}{n}\left( {In} \right)} \right)\\


In=1n1n2(In)In = \frac{1}{n} - \frac{1}{{{n^2}}}\left( {In} \right)\\


In(1+1n2)=1nIn\left( {1 + \frac{1}{{{n^2}}}} \right) = \frac{1}{n}\\


In=1n×n21+n2In = \frac{1}{n} \times \frac{{{n^2}}}{{1 + {n^2}}}\\


In=n1+n2In = \frac{n}{{1 + {n^2}}}


Therefore,


0exsin(nx)dx=n1+n2\int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} = \frac{n}{{1 + {n^2}}}



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