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3 years ago

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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Step-by-step explanation:

From the question, we have:

The random variable is normally distributed according to a standard normal distribution, that is, a normal distribution with $\\ \mu = 0$ and $\\ \sigma = 1$ .
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A z-score is a standardized value, i.e., one that we can obtain using the next formula:

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x is the raw value coming from a normal distribution that we want to standardize.
And we already know that $\\ \mu$ and $\\ \sigma$ are the mean and the standard deviation, respectively, of the normal distribution.

A z-score represents the distance from $\\ \mu$ in standard deviations units. When the value for z is negative, it "tells us" that the raw score is below $\\ \mu$ . Conversely, when the z-score is positive, the standardized raw score, x, is above the mean, $\\ \mu$ .

Solving the question

We already know that $\\ z = -0.84$ or that the standardized value for a raw score, x, is below $\\ \mu$ in 0.84 standard deviations.

The values for probabilities of the standard normal distribution are tabulated in the standard normal table, which is available in Statistics books or on the Internet and is generally in cumulative probabilities from negative infinity, - $\\ \infty$ , to the z-score of interest.

Well, to solve the question, we need to consult the standard normal table for $\\ z = -0.84$ . For this:

Find the cumulative standard normal table.
In the first column of the table, use -0.8 as an entry.
Then, using the first row of the table, find -0.04 (which determines the second decimal place for the z-score.)
The intersection of these two numbers "gives us" the cumulative probability for z or $\\ P(z$ .

Therefore, we obtain $\\ P(z$ for this z-score, or a slightly more than 20% (20.045%) for the "probability that a given score is less than negative 0.84".

This represent the area under the standard normal distribution, $\\ N(0,1)$ , at the left of z = -0.84.

To "draw a sketch of the region", we need to draw a normal distribution (symmetrical bell-shaped distribution), with mean that equals 0 at the middle of the distribution, $\\ \mu = 0$ , and a standard deviation that equals 1, $\\ \sigma = 1$ .

Then, divide the abscissas axis (horizontal axis) into equal parts of one standard deviation from the mean to the left (negative z-scores), and from the mean to the right (positive z-scores).

Find the place where z = -0.84 (i.e, below the mean and near to negative one standard deviation, $\\ -\sigma$ , from it). All the area to the left of this value must be shaded because it represents $\\ P(z$ and that is it.

The below graph shows the shaded area (in blue) for $\\ P(z$ for $\\ N(0,1)$ .

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