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

解

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

θ + 2 sin (θ) + C

解答ステップ

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

三角関数の公式を使用して書き換える

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

拡張

\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 )}

総和規則を適用する:

\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

簡素化

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

= θ + 2 sin (θ)

定数を解答に追加する

= θ + 2 sin (θ) + C