using the  relations
Sum / Difference of Trigonometric Functions Formulas.

   1.  sin A + sin B = 2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ]

   2. sin A - sin B = 2 cos [ (A + B) / 2 ] sin [ (A - B) / 2 ]

   3. cos A + cos B = 2 cos [ (A + B) / 2 ] cos [ (A - B) / 2 ]

    4.  cos A - cos B = - 2 sin [ (A + B) / 2 ] sin [ (A - B) / 2 ]


a) let A= 2x*ex.dx

integrating by parts gives

A = 2x*ex - ln 2 A

so A = 2x*ex / ln(2e) + C   where C is constant

b) Question not clear

c)  cos2x-cos2y/cosx-cosy dx

   = (2 cos2x-1-2cos2y+1)/(cosx-cosy) dx

   = 2 (cosx+cosy) dx


    = 2 (sinx+x.cosy)+ C   where C is constant


d) [sinx*sin(y-x)+sin^2(y/2-x)] dx

  = using relation 4 stated above , we can write as

   = 1/2*( cos(2x-y) - cos y ) + 1/2* (1-cos(2x-y) ) dx

   here cos 2x = 1 - 2sin2x

  so we have finally

  = 1/2 * (1-cosy).dx

  = 1/2 * (1-cosy) + C where C is constant


e) (sin2x+sin5x-sin3x)/(cosx+1-2sin2x) dx


  here here cos 2x = 1 - 2sin2x

   = (sin2x+sin5x-sin3x)/(cosx+cos 4x) dx

   using relations 1 and 3 we have

   =  (-2cox 5x/2. sin x/2+2 sin5x/2 cos5x/2) / (2.cos 5x/2.cos3x/2) dx


  =  ( sin x/2 - sin x/2) / (cos3x/2) dx

  =  (2cox 3x/2. sin x) / (cos3x/2) dx

  =  (2sin x) dx

  =  - 2 cos x  + C where C is constant


f)  (cos8x-cos7x)/(1+2cos5x) dx

   Adding +cos 2x and - cos 2x to numerator gives

         ={(cos8x+cos 2x)-(cos7x+cos 2x)}/(1+2cos5x) dx

       = {(2cos5xcos 3x - (cos7x+cos 2x)}/(1+2cos5x) dx

       again adding +cos 3x and - cos 3x to numerator gives


       = {(2cos5xcos 3x + cos 3x - (cos7x+cos 2x+cos 3x)}/(1+2cos5x) dx


      ={(2cos5xcos 3x + cos 3x - (2 cos 5x.cos 2x+cos 2x)}/(1+2cos5x) dx

      ={(2cos5x+1)cos 3x - (2 cos 5x+1).cos 2x}/(1+2cos5x) dx

     =(cos 3x  - cos 2x) dx

Answer is 1/3*sin 3x - 1/2*sin 2x +C     where C is constant