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∫ dx/1-secx what is the value of the intergal ∑(x i -x - ) 2 solve it

  1. dx/1-secx what is the value of the intergal
∑(xi-x-) solve it

Grade:10

1 Answers

Arun
25763 Points
3 years ago
Dear Trishna
 
  –  ∫ [1 /(secx - 1)] dx = 

rearrange it as: 

 
 – ∫ {1 /[(1/cosx) - 1]} dx = 

 – ∫ {1 /[(1 - cosx) /cosx]} dx = 

 – ∫ [cosx /(1 - cosx)] dx = 

recall the half-angle identity sin²(x/2) = (1 - cosx) /2 

hence 1 - cosx = 2sin²(x/2) 

also, recall the double-angle identity cos(2x) = cos²x - sin²x 

hence cosx = cos²(x/2) - sin²(x/2) 

the integral thus becoming: 

 – ∫ [cosx /(1 - cosx)] dx =  – ∫ [cos²(x/2) - sin²(x/2)] /[2sin²(x/2)] dx = 

break it up into: 

 – ∫ (1/2) [cos²(x/2) /sin²(x/2)] + ∫ (1/2) [sin²(x/2) /sin²(x/2)] dx = 

simplifying, you get: 

 – ∫ (1/2) cot²(x/2) + (1/2) ∫ dx = 

rewrite cot²(x/2) as [csc²(x/2) - 1]: 

 – ∫ (1/2) [csc²(x/2) - 1] + (1/2) ∫ dx = 

break it up into: 

 – ∫ (1/2) csc²(x/2) + (1/2) ∫ dx + (1/2) ∫ dx = 

(adding similar terms together) 

 – ∫ (1/2) csc²(x/2) dx + ∫ dx = 

integrating, you get: 

 cot(x/2) + x + c 

thus, in conclusion: 


∫ [1 /(1 - secx)] dx =  cot(x/2) + x + c 


I hope this helps

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