dx

Find:  ∫ ———————  

               ( 3cosx + 4sinx)

 

                                                                                          2 dz                                                                                           

z = tan (x/2),    x/2 = tan–1(z),    x = 2 tan–1(z),    dx = ———

                                                                                         1+ z2

 

                          x                      2                              2                            2                    2 – 1 – z2       1 – z2

cos x = 2 cos2 (—) – 1 = ————— – 1 = —————— – 1 = ———— – 1 = ————— = ———

                          2               sec2(x/2)               1 + tan2(x/2)               1 + z2                  1 + z2           1 + z2

 

                                                   sin(x/2)                         tan(x/2)           tan(x/2)              2 z

sin x = 2 sin(x/2) cos(x/2) = 2 ———— . cos2(x/2) = ———— = —————— = ———

                                                  cos(x/2)                        sec2(x/2)       1 + tan2(x/2)        1 + z2

 

                               1 – z2             2 z           3 – 3 z2 + 8 z      

3cosx + 4sinx = 3 —–—— + 4 ——— = ——————

 

                               1 + z2          1 + z2             1 + z2         

 

 

               dx                        2 dz               1 + z2   

∫ ———————  = ∫ ——— .——————  

    ( 3cosx + 4sinx)            1+ z2      3 – 3 z2 + 8 z

 

                                                – 2 dz       

                                  = ∫ ——————  

                                              3 z2 – 8 z – 3

 

                                                 – 2 dz       

                                  =  ∫ ——————  

                                           (3z + 1)(z – 3)

 

 

                                           3        dz            1       dz

                                  =∫ [ — . ———  –  — . ——— ]

                                           5     3z + 1         5      z – 3

 

 

                                      1          3dz                 dz

                                  = —[ ∫ ———  –  ∫ ——— ]

                                      5        3z + 1            z – 3

 

                                      1                                                                            

                                  = — [ Ln ( 3z + 1) – Ln ( z – 3) ] + c  .......... (1)

                                      5                                                                        

 

 (Note Ln(a), must a>0)

 

 

                                     1                       x                                x                        

                                  =— [ Ln ( 3 tan — + 1) – Ln (3 – tan — ) ] + c        

                                     5                       2                                2

                     

 OR.    Continue ...... (1)

 

                               sin(x/2)        sin(x/2)     2 cos(x/2)       2 sin(x/2) cos(x/2)           sinx

        z = tan(x/2) =  ——— = ———— . ———— = ———————— = ————   OR see draw

                               cos(x/2)     cos(x/2)     2 cos(x/2)           2 cos2(x/2)              1 + cosx

 

 

 

                                         1                                                                                        sin x             

                                  =   — [ Ln ( 3z + 1) – Ln ( z – 3) ] + c  ( but z = tan(x/2) = ———— ( see draw )

                                         5                                                                                      1 + cosx

 

                                         1                3 sin x                                sin x

                                  =   — [ Ln (  ———— + 1) – Ln ( 3 – ————  ) ] + c     (Note Ln(a), must a>0)  

                                         5              1 + cosx                           1 + cosx

 

 

                                         1              1 + cosx + 3sinx                  3cosx – sinx + 3 

                                  =   — [ Ln (  ———————— ) – Ln ( ——————— ) ] + c  (Ln a/b = Ln a – Ln b)

                                         5                      1 + cosx                               1 + cosx

 

 

                                         1                             

                                  =   — [ Ln ( 3 sin x + cosx + 1) – Ln (  3cosx – sinx + 3 ) ] + c

                                         5