Integration, trig with double substitution

[Azureflames]

Homework Statement

Integral ( cos(x)/(1+cos(x))^.5 dx)

Homework Equations

The Attempt at a Solution

Integral ( cos(x)/(1+cos(x))^.5 dx)

Square it
Integral ( cos(x)^2/(1+cos(x) dx)

Multiply by the Conjugate
Integral ( cos(x)^2/(1+cos(x) * (1 - cos(x))/(1 - cos(x)) dx)

Take the square root and then multiply by sin(x)/sin(x)
Integral ( cos(x)(1-cos(x))^.5/(sin(x) * sin(x)/sin(x) dx )

Convert sin(x)^2 in denom to 1-cos(x)^2 then use u-sub, u= cos(x) du=-sin(x)
Integral -( u(1-u)^.5/(1-u^2) du )

w substitution w= (1-u)^.5 u = 1-w^2 du = -2w
Integral (1-w^2)w^2/(1-(1-2w^2+w^4) dw)

Reduce down to
Integral 2((1-w^2)/(2-w^2) dw )

I have more done but am leaving to go out to dinner now. Will post more of my work later if needed. Thanks in advance.

 

[Dick]

First, you can't 'square' an integral. Second, there is a nice trig identity you can use for the expression inside the square root. Look under half or double angle formulas. Can you find it?

 

[Azureflames]

Okay, after you told me my first step was wrong, I started over. I looked at the half angle double angle stuff and didn't like how it was looking so I tried a similar approach to my first but fixed the first step.

My solution:

Integral ( cos(x)/(1+cos(x))^.5 dx)

Conjugate/trig identities
Integral ( cos(x)/(1+cos(x))^.5 (1-cos(x))^.5/(1-cos(x)^.5) dx)
Integral ( cos(x)(1-cos(x))^.5)/(1-cos(x)^2)^.5 dx)
Integral ( cos(x)(1-cos(x))^.5)/sin(x) dx)

Times sin(x)/sin(x) and trig ident
Integral( sin(x) * cos(x)(1-cos(x))^.5 / (sin(x)^2) dx)
Integral( sin(x) * cos(x)(1-cos(x))^.5 / (1-cos(x)^2 dx)

u-sub u=cos(x) du=-sin(x)
Integral -( u(1-u)^.5/(1-u^2) du)

w-sub w=(1-u)^.5 u=1-w^2 du=-2w dw
Integral 2( (1-w^2)w^2/(1-w^2)^2 dw)

Expanding/reducing
Integral 2((1-w^2)/(2-w^2)

Division
Integral 2((1 - (1/(-w^2+2)) dw)

Partial Fraction Decomposition
2 * Integral (1dw) + .5(2)^.5 * Integral (1/(w-2^.5) dw) - .5(2)^.5 * Integral (1/(w+2^.5) dw)

Integrate

2w + .5(2)^.5*ln(w-2^.5) - .5(2)^.5*ln(w+2^.5) + C

Resubstitute everything...
2(1-cos(x))^.5 + .5(2)^.5*ln((1-cos(x))^.5 - 2^.5) - .5(2)^.5*ln((1-cos(x))^.5 + 2^.5) + C

Fairly certain my steps are correct :D. Thanks for the assistance.

Edit: Really need to figure out formatting lol.

 

[Gib Z]

The Identities that Dick referred to where:

$$cos (2t) = \cos^2 t - \sin^2 t = 2\cos^2 t -1$$.

Both forms come in quite useful in evaluating this integral.

 

[Schrodinger's Dog]

I doubt this is much use but I was having a little play with that on my maths program and it came up with the rather horrendous answer:

? hehe. It looked relatively simple?

Have I expressed your question correctly?

 

[Gib Z]

That expression can be simplified a whole heap!

 

[Schrodinger's Dog]

That expression can be simplified a whole heap!

True but even so it's not very pretty.

 

[tiny-tim]

… combine the ln()s … !

Edit: Really need to figure out formatting lol.

If you type alt-v, it prints √ , which would help a lot

Resubstitute everything...
2(1-cos(x))^.5 + .5(2)^.5*ln((1-cos(x))^.5 - 2^.5) - .5(2)^.5*ln((1-cos(x))^.5 + 2^.5) + C

Fairly certain my steps are correct :D. Thanks for the assistance.

erm … but you haven't finished! The examiners won't like you leaving two ln()s without combining them:

ln(√(1-cos(x)) - √2) - ln(√(1-cos(x)) + √2)

= ln((√1-cos(x)) - √2)/(√(1-cos(x)) + √2))

= … ?
(this can be very much simplified, and I think the result will look familiar to you! )

Okay, after you told me my first step was wrong, I started over. I looked at the half angle double angle stuff and didn't like how it was looking so I tried a similar approach to my first but fixed the first step.

Azureflames, you should do what Dick and Gib Z suggested - they're very wise, and their way only takes about five lines!