<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:
Yes.
Step-by-step explanation:
If you write out the coordinates, you can see that you can do the Vertical Line Test and confirm that if you were to write a function, it would be an actual function. No 2 x-values can be the same y-values. The coordinates all have separate xy values (or the same exact xy values) and you can determine a function.
The values of the given numbers when it is rounded up to the nearest 10 thousands are:
<h3>What is rounding up in mathematics?</h3>
Rounding up can be described as the process that is been used in the mathematics which is been used in the estimation of a particular number in a context.
It should be noted that in rounding the a number up, it is required to look at the next digit at the right hand of the given figures in a case whereby the digit is less than 5,the digit can be rounded down, but in the case whereby the digit is more that 5 then it can be rounded up .
From the given values, we are given the 990,201 and 159,994 and if this were to rounded up to the nearest 10 thousand then we will start from the right hand sides and round down the values less than 5 and round up the values that is more that 5. and their values will be 990000
and 150000.
Read more about rounding up at:
brainly.com/question/28324571
#SPJ1
Answer:
$1,700
Step-by-step explanation:
Given the following :
Period of mortgage = 25 years
Mortgage amount = $130,000
Total interest on mortgage = $380,000
Hence,
Total amount to be paid ( principal + interest) :
($130,000 + $380,000) = $510,000
Total number of months over the 25 years period:
25 × 12 = 300 months
Payment per month:
Total repayment / number of months
= $510,000 / 300
= $1,700 per month