1-1-cot-x-dx-

Question Number 71814 by Aman Arya last updated on 20/Oct/19

\int \frac{1}{1 + \cot x} d x

Commented by mathmax by abdo last updated on 20/Oct/19

l e t I = \int \frac{d x}{1 + c o t a n x} \Rightarrow I = \int \frac{d x}{1 + \frac{c o s x}{s i n x}} = \int \frac{s i n x}{s i n x + c o s x} d x

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e I = \int \frac{\frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

= - 2 \int \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)} d t l e t d e c o m p o s e F (t) = \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)}

t^{2} - 2 t - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2}

F (t) = \frac{t}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}

a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{t_{1}}{(t_{1} - t_{2}) (t_{1}^{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (3 + 2 \sqrt{2} + 1)}

= \frac{2 + \sqrt{2}}{4 (4 + 2 \sqrt{2})} = \frac{2 + \sqrt{2}}{8 (2 + \sqrt{2})} = \frac{1}{8}

b = \frac{t_{2}}{(t_{2} - t_{1}) (t_{2}^{2} + 1)} = \frac{1 - \sqrt{2}}{(- 2 \sqrt{2}) (3 - 2 \sqrt{2} + 1)} = \frac{- 1 + \sqrt{2}}{2 \sqrt{2} (4 - 2 \sqrt{2})}

= \frac{- \sqrt{2} + 2}{8 (2 - \sqrt{2})} = \frac{1}{8}

{l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - a - b = - \frac{1}{4} \Rightarrow

F (t) = \frac{1}{8 (t - t_{1})} + \frac{1}{8 (t - t_{2})} + \frac{- \frac{1}{4} t + d}{t^{2} + 1}

F (0) = 0 = - \frac{1}{8 t_{1}} - \frac{1}{8 t_{2}} + d \Rightarrow d = \frac{1}{8} (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{1}{8} (\frac{t_{1} + t_{2}}{t_{1} t_{2}})

= \frac{1}{8} \frac{2}{- 1} = - \frac{1}{4} \Rightarrow F (t) = \frac{1}{8 (t - t_{1})} + \frac{1}{8 (t - t_{2})} - \frac{1}{4} \frac{t + 1}{t^{2} + 1} \Rightarrow

I = - \frac{1}{4} \int \frac{d t}{t - t_{1}} - \frac{1}{4} \int \frac{d t}{t - t_{2}} + \frac{1}{2} \int \frac{t + 1}{t^{2} + 1} d t

= - \frac{1}{4} l n ∣ t - t_{1} ∣ - \frac{1}{4} l n ∣ t - t_{2} ∣ + \frac{1}{4} l n (t^{2} + 1) + \frac{1}{2} a r c t a n t + c

= - \frac{1}{4} l n ∣ t a n (\frac{x}{2}) - 1 - \sqrt{2} ∣ - \frac{1}{4} l n ∣ t a n (\frac{x}{2}) - 1 + \sqrt{2} ∣

+ \frac{1}{4} l n (1 + {t a n}^{2} (\frac{x}{2})) + \frac{x}{4} + c .

Commented by petrochengula last updated on 21/Oct/19

= \int \frac{1}{1 + \frac{c o s x}{s i n x}} d x

= \int \frac{s i n x}{s i n x + c o s x} d x

= \frac{1}{2} \int \frac{2 s i n x}{s i n x + c o s x} d x

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

= \frac{1}{2} x - \frac{1}{2} l n ∣ s i n x + c o s x ∣ + C

Answered by $@ty@m123 last updated on 20/Oct/19

\int \frac{1}{1 + \frac{\cos x}{\sin x}} d x

\int \frac{\sin x}{\sin x + \cos x} d x

\int \frac{\frac{1}{2} (\sin x + \cos x) - \frac{1}{2} (\cos x - \sin x)}{\sin x + \cos x} d x

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

= \frac{1}{2} x - \frac{1}{2} \ln (\sin x + \cos x) + C