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goldenfox [79]
3 years ago
8

Integrate cosx/sqrt(1+cosx)dx

Mathematics
1 answer:
Step2247 [10]3 years ago
3 0
<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>
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MrRa [10]

1. These angles are complementary, which means that they add up to 90 degrees.

x + 73 = 90

x = 17 degrees

2. These angles are supplementary, which means that they add up to 180 degrees.

x + 118 = 180

x = 62 degrees

Hope this helps!! :)

4 0
3 years ago
O is the centre of the circle. AB is a diameter.
viktelen [127]

Answer:

The value of a is 63° and the value of b is 27°

Step-by-step explanation:

In the given figure

∵ OA, OB, and OC are radii of the circle O

→ The radii of a circle are equal

∴ OA = OB = OC

In Δ AOC

∵ OA = OC

∴ Δ AOC is an isosceles triangle

→ Base angles of the isosceles triangle are equal

∵ ∠OAC and ∠OCA are the base angles of the triangle

∴ m∠OAC = m∠OCA

∵ m∠OAC = a°

∴ m∠OCA = a°

→ The sum of angles in any triangle is 180°

∵ a° + a° + m∠AOC = 180°

∵ m∠AOC = 54°

∴ a° + a° + 54 = 180°

→ Add the like terms

∴ 2a° + 54 = 180°

→ Subtract 54 from both sides

∴ 2a° = 126°

→ Divide both sides by 2

∵ m∠OAC = m∠OCA = 126 ÷ 2

∴ a° = 63

∴ The value of a is 63°

In ΔBOC

∵ OB = OC

∴ Δ BOC is an isosceles triangle

→ Base angles of the isosceles triangle are equal

∵ ∠OBC and ∠OCB are the base angles of the triangle

∴ m∠OBC = m∠OCB

∵ m∠OBC = b°

∴ m∠OCB = b°

∵ ∠AOC is an exterior angle of ΔBOC at the vertex O

→ The measure of the exterior angle equals the sum of the measure

    of the opposite interior angles to this vertex

∵ ∠OBC and ∠OCB are the opposite interior angle of ∠AOC

∴ m∠OBC + m∠OCB = m∠AOC

→ Substitute their measures

∵ b° + b° = 54°

→ Add the like terms

∴ 2b° = 54

→ Divide both sides by 2

∴ b° = 27

∴ The value of b is 27°

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