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Question

Mathematics Question on integral

Find the integrals of the function: cos2x−cos2αcosx−cosα\frac{cos2 x-cos2 α}{cos x - cos α}

Answer

\frac{cos2 x-cos2 α}{cos x - cos α}$$=\frac{ -2 sin (\frac{2x+2α}{2}) sin (\frac{2x-2α}{2}) }{-2sin (\frac{x+α}{2}) sin (\frac{x- α}{2}) } [cos C - cos D = -2sin C+D2\frac{C+D}{2} sin C−D2\frac{C-D}{2}]

=sin(x+α)sin(x−α)sin(x+α2)sin(x−α2)=\frac{sin(x+α)sin(x-α) }{sin(\frac{x+α}{2} )sin(\frac{x-α}{2} )}

=[2sin(x+α2))cos(x+α2))][2sin(x−α2))cos(x−α2))]sin(x+α2))sin(x−α2))=\frac{[2sin(\frac{x+ α}{2}) )cos(\frac{x+ α}{2}) )][2sin(\frac{x- α}{2}) )cos(\frac{x- α}{2}) )]}{sin(\frac{x+ α}{2}) )sin(\frac{x- α}{2}) )}

=4cos(x+α2)cos(x−α2)=4 cos (\frac{x+ α}{2}) cos(\frac{x- α}{2})

=2[cos(x+α2+x−α2)+cosx+α2\-x−α2]=2[cos(\frac{x+ α}{2} +\frac{x- α}{2}) +cos \frac{x+ α}{2} \- \frac{x- α}{2} ]

=2[cos(x)+cosα]=2[cos(x) +cos α]

=2cosx+2cosα=2 cos x+ 2cos α

∴ ∫cos2x−cos2αcosx−cosα\frac{cos2 x-cos2 α}{cos x - cos α} = ∫2cosx+2cosα∫2 cos x+ 2cos α

=2[sinx+xcosα]+C=2[sin x+ x cos α]+C