Question
In=∫π20cosnxcos(nx)dx,nϵN then √I2001:I2002 can be the eccentricity of
Answer: Option D
:
D
In+1=∫π20cosn+1xcos(n+1)xdx=∫π20cosn+1x(cosnxcosx−sinnxsinx)dx∴In+1=In−In+1⇒2In+1=In∴In:In+1=2
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:
D
In+1=∫π20cosn+1xcos(n+1)xdx=∫π20cosn+1x(cosnxcosx−sinnxsinx)dx∴In+1=In−In+1⇒2In+1=In∴In:In+1=2
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