integral: [Int] ( 1 + cos x ) ^ 1/2 dx: tried 'by parts' but

[peblez]

[Int] ( 1 + cos x ) ^ 1/2 dx

I tried using integration by parts but then i get stuck after that procedure.

 

[soroban]

Re: Hard integral

Hello, peblez!

\(\displaystyle \int (1 + \cos x)^{\frac{1}{2}}dx\)

Fastest method

\(\displaystyle \text{Use the identity: }\;1 + \cos x \:=\:2\cos^2\left(\frac{x}{2}\right)\)


Impressive method:

\(\displaystyle \sqrt{1 + \cos x} \;=\;\sqrt{\frac{1 + \cos x}{1}\cdot\frac{1-\cos x}{1-\cos x}} \;=\;\sqrt{\frac{1-\cos^2\!x}{1-\cos x}} \;=\;\sqrt{\frac{\sin^2\!x}{1-\cos x}} \;=\;\frac{\sin x}{\sqrt{1-\cos x}}\)

\(\displaystyle \text{And we have: }\;\int(1-\cos x)^{-\frac{1}{2}}(\sin x\,dx)\)

\(\displaystyle \text{Now let: }\:u \:=\:1-\cos x\)

 

[galactus]

Re: Hard integral

Soroban's metods are certainly better than this, but sometimes when confronted with a tough integral you can use the substitution

\(\displaystyle x=2tan^{-1}(u), \;\ dx=\frac{2}{u^{2}+1}du, \;\ u=tan(\frac{x}{2})\)

Making the subs gives the integral:

\(\displaystyle 2\sqrt{2}\int(u^{2}+1)^{\frac{-3}{2}}du\)

This can be solved by using a trig sub: \(\displaystyle u=tan({\theta}), \;\ du=sec^{2}({\theta})d{\theta}\)

Just throwing it out there.