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TEA [102]
3 years ago
6

Please answer question now fast

Mathematics
1 answer:
german3 years ago
7 0

Answer:

its a

Step-by-step explanation:

on edge

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Integrate cosx/sqrt(1+cosx)dx
Step2247 [10]
<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>
3 0
3 years ago
Which method would you use to prove that the two triangles are congruent?
Alona [7]

Answer:- AAS postulate


Explanation:-

  • AAS postulate tells that if two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

Given:- One angle and one side of a triangle is equal to the one angle and one side of the other triangle.

We see there is one more pair of equal angles as they are vertically opposite angles . [See the attachment]

⇒ there is a triangle where two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

⇒ The triangles are congruent [ by ASA postulate]


6 0
4 years ago
Geralds baby brother spends 7/8 ofhis day sleeping. How many hours does his baby brother sleep
Effectus [21]
His baby brother sleeps 21 hours.
There's a total of 24 hours in a day so you can divide by 8. That means 1/8 of the day is equal to 3 hours, so 7 times 3 hours equals 21 hours
6 0
4 years ago
The product of (-2)(-5)(-7)
Effectus [21]
(-2)(-5)(-7)
(10)(-7)
-70
The product is -70
3 0
3 years ago
Read 2 more answers
The length of a rectangle is 1 less than twice the width. The area of the rectangle is 28 square feet. The equation 2w²-w-28 , w
dybincka [34]
The answer should be B
4 0
3 years ago
Read 2 more answers
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