# Integration, trig with double substitution

1. Homework Statement

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

2. Homework Equations

3. 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.

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
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?

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
Homework Helper
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.

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
Homework Helper
That expression can be simplified a whole heap!

That expression can be simplified a whole heap!
True but even so it's not very pretty.

tiny-tim
Homework Helper
… 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!