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goldenfox [79]
3 years ago
8

Integrate cosx/sqrt(1+cosx)dx

Mathematics
1 answer:
Step2247 [10]3 years ago
3 0
<span>Take the integral: integral (cos(x))/sqrt(cos(x)+1) dx
For the integrand (cos(x))/sqrt(1+cos(x)), substitute u = 1+cos(x) and du = -sin(x) dx: = integral (u-1)/(sqrt(2-u) u) du
For the integrand (-1+u)/(sqrt(2-u) u), substitute s = sqrt(2-u) and ds = -1/(2 sqrt(2-u)) du: = integral -(2 (1-s^2))/(2-s^2) ds
Factor out constants: = -2 integral (1-s^2)/(2-s^2) ds
For the integrand (1-s^2)/(2-s^2), cancel common terms in the numerator and denominator: = -2 integral (s^2-1)/(s^2-2) ds
For the integrand (-1+s^2)/(-2+s^2), do long division: = -2 integral (1/(s^2-2)+1) ds
Integrate the sum term by term: = -2 integral 1/(s^2-2) ds-2 integral 1 ds
Factor -2 from the denominator: = -2 integral -1/(2 (1-s^2/2)) ds-2 integral 1 ds
Factor out constants: = integral 1/(1-s^2/2) ds-2 integral 1 ds
For the integrand 1/(1-s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds: = sqrt(2) integral 1/(1-p^2) dp-2 integral 1 ds
The integral of 1/(1-p^2) is tanh^(-1)(p): = sqrt(2) tanh^(-1)(p)-2 integral 1 ds
The integral of 1 is s: = sqrt(2) tanh^(-1)(p)-2 s+constant
Substitute back for p = s/sqrt(2): = sqrt(2) tanh^(-1)(s/sqrt(2))-2 s+constant
Substitute back for s = sqrt(2-u): = sqrt(2) tanh^(-1)(sqrt(1-u/2))-2 sqrt(2-u)+constant
Substitute back for u = 1+cos(x): = sqrt(2) tanh^(-1)(sqrt(sin^2(x/2)))-2 sqrt(1-cos(x))+constant
Factor the answer a different way: = sqrt(1-cos(x)) (csc(x/2) tanh^(-1)(sin(x/2))-2)+constant
Which is equivalent for restricted x values to:
Answer: | | = (2 cos(x/2) (2 sin(x/2)+log(cos(x/4)-sin(x/4))-log(sin(x/4)+cos(x/4))))/sqrt(cos(x)+1)+constant</span>
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The range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

Step-by-step explanation:

We are given that the lengths of pregnancies in a small rural village are normally distributed with a mean of 262 days and a standard deviation of 17 days.

Let X = <u><em>lengths of pregnancies in a small rural village</em></u>

SO, X ~ Normal(\mu=262,\sigma^{2} = 17^{2})

Here, \mu = population mean = 262 days

         \sigma = standard deviation = 17 days

<u>Now, the 68-95-99.7 rule states that;</u>

  • 68% of the data values lies within one standard deviation points.
  • 95% of the data values lies within two standard deviation points.
  • 99.7% of the data values lies within three standard deviation points.

So, middle 68% of most pregnancies is represented through the range of within one standard deviation points, that is;

[ \mu -\sigma , \mu + \sigma ]  =  [262 - 17 , 262 + 17]

                          =  [245 days , 279 days]

Hence, the range in which we can expect to find the middle 68% of most pregnancies is [245 days , 279 days].

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