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∫ 1/ (cos3x – cosx ) dx , please say how to simplyfy, thank you.
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
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
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