Friday, 13 December 2013

calculus - Evaluate : intfracpi20fraccos2x,dxcos2x+4sin2x

Evaluate: π20cos2xdxcos2x+4sin2x




First approach :



π20cos2xdxcos2x+4(1cos2x)



=π20cos2xdx43cos2x



=π2013{43cos2x443cos2x}dx



=π2013{1443cos2x}dx




=π20131dxπ20134sec2x4sec2x3dx



=π20131dxπ20134sec2x4(1+tan2x)3dx



Now I can easily put tanx=t and I get sec2xdx=dt



Second approach :




π20cos2xdxcos2x+4sin2x



Dividing numerator and denominator by cos2x we get :



=π20dx1+4tan2x



=π20dx(4){14+tan2x}



=π20dx(4){{12}2+(tanx)2}




Can we apply this formula of integral here :



1a2+x2dx=1atan1xa



I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...

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