What is the integral of (cos(θ)-cos(2θ))/(1-cos(θ)) ?

The integral of (cos(θ)-cos(2θ))/(1-cos(θ)) is θ+2sin(θ)+C

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integral of (cos(θ)-cos(2θ))/(1-cos(θ))

\int \frac{cos ( θ ) - cos ( 2 θ )}{1 - cos ( θ )} d θ

θ + 2 sin (θ) + C

Solution steps

\int \frac{cos ( θ ) - cos ( 2 θ )}{1 - cos ( θ )} d θ

Rewrite using trig identities

= \int \frac{- cos ^{2} ( θ ) + sin ^{2} ( θ ) + cos ( θ )}{1 - cos ( θ )} d θ

Expand

\frac{- cos ^{2} ( θ ) + sin ^{2} ( θ ) + cos ( θ )}{1 - cos ( θ )} : - \frac{cos ^{2} ( θ )}{1 - cos ( θ )} + \frac{sin ^{2} ( θ )}{1 - cos ( θ )} + \frac{cos ( θ )}{1 - cos ( θ )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{cos ^{2} ( θ )}{1 - cos ( θ )} d θ + \int \frac{sin ^{2} ( θ )}{1 - cos ( θ )} d θ + \int \frac{cos ( θ )}{1 - cos ( θ )} d θ

\int \frac{cos ^{2} ( θ )}{1 - cos ( θ )} d θ = - cot (\frac{θ}{2}) - sin (θ) - θ

\int \frac{sin ^{2} ( θ )}{1 - cos ( θ )} d θ = sin (θ) + θ

\int \frac{cos ( θ )}{1 - cos ( θ )} d θ = - cot (\frac{θ}{2}) - θ

= - (- cot (\frac{θ}{2}) - sin (θ) - θ) + sin (θ) + θ - cot (\frac{θ}{2}) - θ

Simplify

- (- cot (\frac{θ}{2}) - sin (θ) - θ) + sin (θ) + θ - cot (\frac{θ}{2}) - θ : θ + 2 sin (θ)

= θ + 2 sin (θ)

Add a constant to the solution

= θ + 2 sin (θ) + C

x (ln (x))

\frac{\partial}{\partial x} (x + y - 8)

f (x) = (x^{2} + 5) (8 - x)

x \to 1 lim (4 x^{2} + 2)

\frac{4}{x} - 1

What is the integral of (cos(θ)-cos(2θ))/(1-cos(θ)) ?
The integral of (cos(θ)-cos(2θ))/(1-cos(θ)) is θ+2sin(θ)+C