Q: Compute: ∫√(2 – 2cosx) dx
A: This is tricky.. Follow along each step with pen and paper:
Start with the following u-substitution:
u = 2 – 2cosx
du = 2sinx dx
Solve for dx:
du / (2sinx) = dx
I need the left side of the above equation to be all in u’s and with no x’s…
Therefore, follow my logic.. We know that:
u = 2 – 2cosx
u – 2 = -2cosx
(-1/2)(u – 2) = cosx
(2 – u) / 2 = cosx
So, draw a triangle and estiblish x as one of the angles. We know that cosine is adjacent/hypotenuse… So the adjacent side = 2 – u and the hypotenuse = 2
Use the Pythagorean Theorem to solve for the missing side (I will call it b):
a² + b² = c²
(2 – u)² + b² = 2²
4 – 4u + u² + b² = 4
b² = 4u – u²
b = √(4u – u²)
So, sinx = opposite side / hypotenuse = b / 2 = √(4u – u²) / 2
Remember:
du / (2sinx) = dx
Now we know that sinx = √(4u – u²) / 2, so we can substitute that into the blue equation:
du / (2sinx) = dx
du / (2(√(4u – u²) / 2)) = dx
Now simplify:
du / (2(√(4u – u²) / 2)) = dx
du / √(4u – u²) = dx
OK! Back to the original problem:
∫√(2 – 2cosx) dx
Substitute in “u” and “du” like so:
∫√(2 – 2cosx) dx = ∫√(u) du / √(4u – u²)
Now, clean up:
∫√(u)/√(4u – u²) du
Now, I am going to factor the bottom and separate the square-root like so:
∫√(u)/√(u(4 – u)) du = ∫√(u)/(√(u)*√(4 – u)) du
We can cancel a √u out of the top and bottom to get:
∫ 1 / √(4 – u) du
Now, another substitution:
s = 4 – u
ds = – du OR -ds = du
Now, put our s’s in to get:
∫ 1 / √(4 – u) du = ∫ – 1 / √(s) ds
Re-write:
∫ – 1 / √(s) ds = ∫ – s-1/2 ds
Now integrate to get:
∫ – s-1/2 ds = -2s1/2+ C
s = 4 – u
So, -2s1/2+ C = -2(4 – u)1/2+ C
And, u = 2 – 2cosx, so:
-2(4 – u)1/2+ C = -2(4 – (2 – 2cosx)1/2+ C = -2(2 + 2cosx)1/2+ C
(The solution on WolframAlpha shows that I am off by a negative… I cannot for the life of me figure out how… Anyone know??)
here are the links to the work I did
1. http://twitpic.com/mum3l
2. http://twitpic.com/mum3t
3. http://twitpic.com/mum3k
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