tan x = sin x/cos x
cos³ x *tan²x - cos x = cos³x *(sin²x/ cos² x) - cos= cosx*sin²x - cos x =
= cos x(sin² x - 1) = - cos x(1 - sin²x)= - cosx*cos²x = - cos³ x
140 ÷ 9= _____. ____ < or > than 15.
140 ÷ 9 is 15.55555556. 15.55555556>15. they need 16 vans.
<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:
skiing- 8/1
hiking- 6/1
Step-by-step explanation:
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Answer:
Kahoot: The Emoji Quiz
Meet: wff-qjue-fja
DONT WORRY IM NOT A OLD MAN
PLSDHQGH.
Step-by-step explanation:
