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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
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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
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