Question

Friends I am preparing for nest help me in this question

find the integration of ∫ 1dx / [ 3sinx + 4cosx + 7 ]

Answer 1

 hello pari  :-) 

our question is 

∫ 1dx / [ 3sinx + 4cosx + 7 ] 

∫ dx / [ 3 ( 2tanx/2 / 1+tan²(x/2) )    +   4 (1-tan²(x/2)/(1+tan²(x/2) + 7 ]   as we know that  

sin x = 2tan(x/2)/(1-tan²(x/2)        and cos x = [1-tan²(x/2) ] / [1+ tan²(x/2) ]

we get    ∫  (1+tan²(x/2) dx /[ 3 tan (x/2 )  +4 ( 1-tan² (x/2)    + 7 ( 1+tan²(x/2) ] 

and we know that 1+tan²(x/2) = > sec ²(x/2) 

∫ sec²(x/2) dx /  /[ 3 tan (x/2 )  +4 ( 1-tan² (x/2)    + 7 sec ²(x/2) ] 

let tan (x/2) = > t 

1/2 sec²(x/2)dx = > dt which means we get 

∫ dt/[ 3 ×2t +4 ( 1 - t²)  + 7 ( 1+ t²) ]   as we know that sec²(x/2) = > 1+ tan²(x/2)

now we will reduce this term by solving inner part pari 

∫ dt /[ 3×2t +4  - 4t²  + 7+ 7t²) ]

∫ dt /  [ 6t ² + 3t  + 11 ] 

∫dt /  [ 3 [ t² +2t ] +11 ] 

∫dt /[ 3[ (t+1)² -1] +11 ] 

∫dt / 3(t+1)² +8]    and we know its integration is 

1/√24  ( tan inverse ( √3  ×(t+1) / √8 ) is our required answer 

Hope this will help you ! 

@ gopal khandelwal b-tech from IIT ROORKEY