∫ √[1 + cos(x)] dx : how do one integrate this?

Question

∫ √[1 + cos(x)] dx

this is not in the forms

1/(1 + x²), or 1/√[1 - x²]

so that I could use the trig identities I know

now I'm confused

pleas do show me the steps. thank you

Answer 1

The point is to use identities you don't know :-). Here's a link to a huge bunch of them.

http://en.wikipedia.org/wiki/Trigonometr...

∫ √(1 + cos x) dx = I

∫ √((1 + cos x)/2) dx = l/√2

∫ cos (x/2) dx = I/√2

x/2 = u

dx/2 = du

dx = 2 du

2 ∫ cos u du = I/√2

2 sin u + C = I/√2

2√2 sin (x/2) + C = l

ANS: 2√2 sin (x/2) + C = I

Answer 2

∫ √[1 + cos(x)] dx  = ????

1+cos2(x/2) =2cos^2 (x/2)  

(since we know 1+cos2x = 2cos^x  

so = ∫ √[1 + cos(x)] dx  = √2  ÃƒÂ¢Ã‚ˆÂ« cos^2(x/2)] dx  

= √2 [ sin(x/2)]/2  =  (sin(x/2)) / √2   answer

good luck

Answer 3

you always post interesting questions...!

∫ √(1 + cosx) dx =

try the substitution cosx = u

thus

x = arccos u →

dx = [- 1/√(1 - u²)] du

thus, substituting, you get:

∫ √(1 + cosx) dx = ∫ √(1 + u) [- 1/√(1 - u²)] du =

- ∫ [√(1 + u) /√(1 - u²)] du =

factor the radicand in the denominator as a difference of squares:

- ∫ {√(1 + u) /√[(1 + u)(1 - u)]} du =

- ∫ {√(1 + u) /[√(1 + u)√(1 - u)]} du =

√(1 + u) canceling out,

- ∫ [ 1/√(1 - u)] du =

that is the same as:

∫ [ (- du) / √(1 - u)]  =

∫ [ d(1 - u) / √(1 - u)]  =

∫ [(1 - u)^(-1/2)]  d(1 - u) =

having both the function (1 - u) and its differential,

{(1 - u)^[(-1/2)+1]}/[(-1/2)+1] + C =

[(1 - u)^(1/2)]/(1/2) + C =

2√(1 - u) + C

that is, substituting back u = cosx:  

∫ √(1 + cosx) dx = 2√(1 - cosx) + C

I hope it helps..

(I'm pretty sure that this answer will get a thumb down....)

Bye!