<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>
We have a three unknown, 4 equation homogeneous system. These always have at least (0,0,0) as a solution. Let's write the equations, one column at a time.
1a + 0b + 0c = 0
-1a + 1b +0c = 0
0a - 1b + c = 0
0a + 0b + -1 c = 0
We could do row reduction but these are easy enough not to bother.
Equation 1 says
a = 0
Equation 4 says
c = 0
Substituting in the two remaining,
-1(0) + 1b + 0c = 0
b = 0
0(0) - 1b + 0 = 0
b = 0
The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)
Angle B is also 54° because they are vertically opposite angles.
Now, you do 54+54=108 and then 360-108=252. (The angles should always add up to 360, so remember that) Because angles C and D are also vertically corresponding angles, they add up to 252, and each measure 126°
B) 405 m
tan(5) = x/4629
4629tan(5) = 404.9
Round up
x=405 m
First calculate
.
We take positive value of sinθ because 0° < θ < 90°
Now we could calculate sin2θ, cos2θ and tan2θ:
