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roshni jain Grade: 12
        

∫1⁄cos3x-cosx dx

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

  dx/cos3x-cosx =dx/4cos^3x-4cosx   by using identity   cos3x =4cos^3x - 3cosx


=  dx/4.cosx(cos^2x-1)


=cosxdx/4cos^x(cos^x-1)


=-cosxdx/4sin^2x(1-sin^x)


 now putting sinx=t,cosxdx=dt


  =-dt/4(1-t^2)t^2


  =1/4 {-dt/t^2  -dt/1-t^2 }


  =1/4 .{1/t + log1-t/1+t } +c


 =1/4 .(cosecx + log1-sinx/1+sinx } +c


 

6 years ago
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