\t∫dx/1-secx what is the value of the intergal∑(xi-x-)2solve it
Trishna mondal , 6 Years ago
Grade 10
Integral Calculus> ∫ dx/1-secx what is the value of the inte...
1 Answers
Arun
Last Activity: 6 Years ago
Dear Trishna
– ∫ [1 /(secx - 1)] dx =
rearrange it as:
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
– ∫ {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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