cos-sin-cos-d-

Question Number 102183 by bemath last updated on 07/Jul/20

\int \frac{\cos θ}{\sin θ - \cos θ} d θ ?

Answered by Dwaipayan Shikari last updated on 07/Jul/20

\frac{1}{2} l o g (s i n θ - c o s θ) - \frac{θ}{2} + C

I = \int \frac{c o s θ}{s i n θ - c o s θ} d θ

I = \int \frac{s i n θ}{s i n θ - c o s θ} - 1 d θ

2 I = \int \frac{s i n θ + c o s θ}{s i n θ - c o s θ} - 1 d θ

I = \frac{1}{2} l o g (s i n θ - c o s θ) - \frac{θ}{2} + C o n s t a n t

Answered by bobhans last updated on 07/Jul/20

\frac{\cos θ}{\sin θ - \cos θ} \times \frac{\sin θ + \cos θ}{\sin θ + \cos θ} = \frac{\frac{1}{2} \sin 2 θ + \cos^{2} θ}{- \cos 2 θ}

\frac{1}{4} \int \frac{d (\cos 2 θ)}{\cos 2 θ} - \int \frac{\cos^{2} θ}{\cos 2 θ} d θ =

\frac{1}{4} \ln ∣ \cos 2 θ ∣ - \int \frac{\frac{1}{2} + \frac{1}{2} \cos 2 θ}{\cos 2 θ} d θ =

\frac{1}{4} \ln ∣ \cos 2 θ ∣ - \frac{1}{2} \int \sec 2 θ - \frac{1}{2} θ + c =

\frac{1}{4} \ln ∣ \cos 2 θ ∣ - \frac{1}{4} \ln ∣ \sec 2 θ + \tan 2 θ ∣ - \frac{1}{2} θ + c

Answered by 1549442205 last updated on 07/Jul/20

Putting t = \tan \frac{θ}{2} \Rightarrow dt = \frac{1}{2} (1 + \tan^{2} \frac{θ}{2}) d θ = \frac{1}{2} (1 + t^{2}) d θ

\cos θ = \frac{1 - t^{2}}{1 + t^{2}}, \sin θ = \frac{2 t}{1 + t^{2}}, d θ = \frac{2 dt}{1 + t^{2}}

F = 2 \int \frac{1 - t^{2}}{(t^{2} + 2 t - 1) (1 + t^{2})} dt == - \int \frac{t + 1}{t^{2} + 1} dt + \int \frac{t + 1}{t^{2} + 2 t - 1} dt

= - \int \frac{dt}{1 + t^{2}} - \frac{1}{2} \int \frac{d (t^{2} + 1)}{t^{2} + 1} + \frac{1}{2} \int \frac{d (t^{2} + 2 t - 1)}{t^{2} + 2 t - 1}

= - \arctan t - \frac{1}{2} \ln (1 + t^{2}) + \frac{1}{2} \ln ∣ t^{2} + 2 t - 1 ∣

\frac{- θ}{2} - \frac{1}{2} \ln (\tan^{2} \frac{θ}{2} + 1) + \frac{1}{2} \ln ∣ \tan^{2} \frac{θ}{2} + 2 \tan \frac{θ}{2} - 1 ∣ + C

Answered by mathmax by abdo last updated on 08/Jul/20

I = \int \frac{\cos θ}{sn θ - \cos θ} d θ \Rightarrow I = \int \frac{d θ}{\tan θ - 1} changement \tan θ = x give

I = \int \frac{dx}{(1 + x^{2}) (x - 1)} let decompose F (x) = \frac{1}{(x - 1) (x^{2} + 1)} \Rightarrow

F (x) = \frac{a}{x - 1} + \frac{bx + c}{x^{2} + 1}

wehave a = \frac{1}{2}, \lim_{x \to + \infty} xF (x) = 0 = a + b \Rightarrow b = - \frac{1}{2}

F (0) = - 1 = - a + c \Rightarrow c = a - 1 = - \frac{1}{2} \Rightarrow F (x) = \frac{1}{2 (x - 1)} + \frac{- \frac{1}{2} x - \frac{1}{2}}{x^{2} + 1} \Rightarrow

I = \frac{1}{2} \int \frac{dx}{x - 1} - \frac{1}{2} \int \frac{x + 1}{x^{2} + 1} dx = \frac{1}{2} \ln ∣ x - 1 ∣ - \frac{1}{4} \ln (x^{2} + 1) - \frac{1}{2} arctanx + c

= \frac{1}{2} \ln ∣ \tan θ - 1 ∣ - \frac{1}{4} \ln (1 + \tan^{2} θ) - \frac{θ}{2} + c .