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 = -2s^1/2+ C

s = 4 – u

So, -2s^1/2+ 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??)