8 1/2 . answered from desmos calculator
<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>
The answer is <span>10 units
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The diameter of the cylinder (d) is the twice of its radius (r): d = 2r
The volume (V) of the cylinder with radius r and height h is:
V = π r² h
We have:
d = 2r
r = ?
h = 7
π = 3.14
V = 175π
175π = π * r² * 7
175 = r² * 7
r² = 175/7
r² = 25
r = √25
r = 5
d = 2r
d = 2 * 5
d = 10
Answer: 5 to the 2 power x4+100 divide by 2
Step-by-step explanation:
Answer:
2 units
Step-by-step explanation:
Given data
The points are (2, 3) and (2, 5)
x1= 2
y1= 3
x2= 2
y2= 5
substitute into the distance between two points formula
d=√((x_2-x_1)²+(y_2-y_1)²)
d=√((2-2)²+(5-3)²)
d=√(2)²)
d=√4
d= 2 units
Hence the length is 2 units
