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anonymous
 3 years ago
integrate cosx/sqrt(1+cosx)dx
anonymous
 3 years ago
integrate cosx/sqrt(1+cosx)dx

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ybarrap
 3 years ago
Best ResponseYou've already chosen the best response.0Take 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 (u1)/(sqrt(2u) u) du For the integrand (1+u)/(sqrt(2u) u), substitute s = sqrt(2u) and ds = 1/(2 sqrt(2u)) du: = integral (2 (1s^2))/(2s^2) ds Factor out constants: = 2 integral (1s^2)/(2s^2) ds For the integrand (1s^2)/(2s^2), cancel common terms in the numerator and denominator: = 2 integral (s^21)/(s^22) ds For the integrand (1+s^2)/(2+s^2), do long division: = 2 integral (1/(s^22)+1) ds Integrate the sum term by term: = 2 integral 1/(s^22) ds2 integral 1 ds Factor 2 from the denominator: = 2 integral 1/(2 (1s^2/2)) ds2 integral 1 ds Factor out constants: = integral 1/(1s^2/2) ds2 integral 1 ds For the integrand 1/(1s^2/2), substitute p = s/sqrt(2) and dp = 1/sqrt(2) ds: = sqrt(2) integral 1/(1p^2) dp2 integral 1 ds The integral of 1/(1p^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(2u): = sqrt(2) tanh^(1)(sqrt(1u/2))2 sqrt(2u)+constant Substitute back for u = 1+cos(x): = sqrt(2) tanh^(1)(sqrt(sin^2(x/2)))2 sqrt(1cos(x))+constant Factor the answer a different way: = sqrt(1cos(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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ybarrap (u1)/(sqrt(2u) u) du how do you get (sqrt(2u)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ \sqrt{2} } \int\limits_{}^{}\frac{ cosx }{ \sqrt{\frac{ 1+cosx }{ 2 }} }dx\] \[1/\sqrt{2}\int\limits_{}^{}\frac{ \cos(x) }{ \cos(x/2) }dx \rightarrow 1/\sqrt{2} \int\limits_{}^{}\frac{ 2\cos^2(x/2)1 }{ \cos(x/2) }dx\] \[1/\sqrt{2}\int\limits_{}^{}2\cos(x/2)\sec(x/2)dx\] From here I think you can handle the rest..
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