<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>
Answer:
%82.5
Step-by-step explanation:
- The final exam of a particular class makes up 40% of the final grade
- Moe is failing the class with an average (arithmetic mean) of 45% just before taking the final exam.
From point 1 we know that Moe´s grade just before taking the final exam represents 60% of the final grade. Then, using the information in the point 2 we can compute Moe´s final grade as follows:
,
where FG is Moe´s Final Grade and FE is Moe´s final exam grade. Then,
.
So, in order to receive the passing grade average of 60% for the class Moe needs to obtain in his exam:
That is, he need al least %82.5 to obtain a passing grade.
The necessary formulas to get the area of a circle’s segment are the area of the circle’s sector and the triangle it is bounded by.
Hence the formula for the segment’s area is:Area of Segment = Area of sector – area of the triangle.
The formula of the area of a circle’s sector is:Area of Sector =
x (radius of circle)²
While the formula for the area of a triangle is:Area of Triangle =
x base of triangle x height of triangle
There really is no ‘best stock’ to invest in. But some good stocks to invest in are Amazon, Disney, and Apple. All are companies that have had tremendous growth and appear to still be growing. You can also go onto Yahoo finance, they will tell you whether they think the stock is over priced and you can also see the growth pattern over time.
Hope this helps! Please make me the brainliest, it’s not necessary but appreciated, I put a lot of effort and research into my answers. Have a good day, stay safe and stay healthy.
Consider, pls, this option.
Please, change the design according to local requirements.
