Verified answer

dx / (1 - cos(x)) =>

(1 + cos(x)) * dx / ((1 - cos(x)) * (1 + cos(x))) =>

(1 + cos(x)) * dx / (1 - cos(x)^2) =>

(1 + cos(x)) * dx / sin(x)^2 =>

dx / sin(x)^2 + cos(x) * dx / sin(x)^2 =>

csc(x)^2 * dx + cos(x) * dx / sin(x)^2

The integral of csc(x)^2 * dx is -cot(x)

To integrate cos(x) * dx / sin(x)^2, we use u-substitution

u = sin(x)

du = cos(x) * dx

du / u^2

Integrate

-1/u + C =>

-1/sin(x) + C =>

-csc(x) + C

Now we have:

-cot(x) - csc(x) + C

This requires some knowledge of double angle formulas. Recall that cos(2x) = 1 - 2sin^2(x). This meant that 1 - cos(2x) = 2sin^2(x). If we let x = 1/2x, then we can have 1 - cos(x) = 2sin^2(x/2)

Therefore it's simplified to integrating 1/2sin^2(x/2) dx.

From here, you might know that the derivative of the cotangent function cot(x) is -1/sin^2(x).

This means the derivative of cot(x/2) = -1/2*sin^2(x/2) which is similar but in fact the negative of the above integral. Hence d/dx(-cot(x/2)) gives the above integral 1/2*sin^2(x/2).

So the integral can be equated to -cot(x/2) + c. This method is a more on the intuition side rather than calculations.

a million/(a million+cosx) = (a million-cosx)/[(a million+cosx).(a million-cosx)] = (a million-cosx)/(a million-cos²x) = = a million/(a million-cos²x) - cosx/(a million-cos²x) = a million/sin²x -cosx/sin²x So, Int[dx/(a million+cos²x)] = Int(dx/sin²x) - Int(cosx/sin²x) enable us to remedy the 1st term of the sum: Int(dx/sin²x), making u = cosx/sinx we've: du = (-sin²x - cos²x)dx/sin²x or: dx = -sin²x.du Int(dx/sin²x) = Int(-sin²x.du/sin²x) = -Int(du) = u+C = cosx/senx +C Now enable us to remedy the 2d term: -Int(cosxdx/sin²x); if we make u = sinx, du = -cosx dx -int(cosxdx/sin²x) = -int(-du/u²) = int(du/u²) = -a million/u +C so, the end results of the mixing is: Int(dx/(a million+cosx) = cosx/sinx - a million/u +C = cotg(x) -a million/u +C

∫1/(1 - cos(x)) dx

∫(1 + cos(x))/(1 - cos²(x)) dx

∫(1 + cos(x))/sin²(x) dx

∫(csc²(x) + cos(x)/sin²(x)) dx

= -cot(x) - csc(x) + C