find ∫{(3+4cosx)/(4+3cosx)²}dx.

plz give solution in detail.

Cos2x= 1-Tan^2 x/ 1+Tan^2 x { Cosx= 1-Tan square x whole divided by 1+ Tan square x}

using this formula,the function will reduce to f(t)= 2(7-t^2)/(7+t^2)^2, where t= Tanx/2

Now u can use simple algebraic operations to solve it

HOPE IT HELPS

ALL THE BEST

PLZZZ APPROVE MY ANSWER IF U LIKE IT

       f(x)= 3+4cosx/(4+3cosx)^2

      putting cosx=1-tan^2(x/2) /1+tan^2(x/2)       

                                                                                               [from cos2x=1-tan^2(x)/1+tan^2(x)]

 f(x) becomes 7-tan^2(x/2) .sex^2(x/2)/7+tan^2(x/2)

       I=  int [7-tan^2(x/2) . sex^2(x/2)/7+tan^2(x/2)]

        put tanx/2 =t

     after differentiating we get

               sex^2(x/2)dx=dt

       putting in previous eq we get

     I=  int 7-t^2/7+t^2 or -(t^2-7)/(t^2+7)

      =  int  t^2+7-14/t^2+7

       =int {  (1)  -(14/t^2+7)  }   after separating                                                      int 1/x^2+a^2 =1/ataninverse(x/a)

        =       t  -  14/root7taninverse(t/root7) + c

     again substituting value of t we get

    I =tanx/2  -  2root7(taninverse{tan(x/2))/root7} +c