Integrate cosx/sqrt(1+cosx)dx

1 answer:

3 0

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

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Suppose there is a 13.9 % probability that a randomly selected person aged 40 years or older is a jogger. In addition, there is

Blababa [14]

Answer:

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

It would be unusual to randomly select a person aged 40 years or older who is male and jogs.

Step-by-step explanation:

We have these following probabilities.

A 13.9% probability that a randomly selected person aged 40 years or older is a jogger, so $P(A) = 0.13$ .

In addition, there is a 15.6% probability that a randomly selected person aged 40 years or older is male comma given that he or she jogs. I am going to say that P(B) is the probability that is a male. $P(B/A)$ is the probability that the person is a male, given that he/she jogs. So $P(B/A) = 0.156$

The Bayes theorem states that:

$P(B/A) = \frac{P(A \cap B)}{P(A)}$

In which $P(A \cap B)$ is the probability that the person does both thigs, so, in this problem, the probability that a randomly selected person aged 40 years or older is male and jogs.

$P(A \cap B) = P(A).P(B/A) = 0.156*0.139 = 0.217$

There is a 2.17% probability that a randomly selected person aged 40 years or older is male and jogs.

A probability is unusual when it is smaller than 5%.

So it would be unusual to randomly select a person aged 40 years or older who is male and jogs.

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3 years ago

Number of stickers is less than 1145 but greater than 1100. How many stickers?

dimaraw [331]

45 stickers because 1145 - 1100 = 45.

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2 years ago

In testing a certain kind of missile, target accuracy is measured by the average distance X at which the missile explodes. The d

3241004551 [841]

Answer:

The answer is c) 761.0

Step-by-step explanation:

Mathematical hope (also known as hope, expected value, population means or simply means) expresses the average value of a random phenomenon and is denoted as E (x). Hope is the sum of the product of the probability of each event by the value of that event. It is then defined as shown in the image, Where x is the value of the event, P the probability of its occurrence, "i" the period in which said event occurs and N the total number of periods or observations.

The variance of a random variable provides an idea of the dispersion of the random variable with respect to its hope. It is then defined as shown in the image.

Then you first calculate E [x] and E [ $x^{2}$ ], and then be able to calculate the variance.