∫(1/(1+cot x))dx


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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
$let I=∫ (dx/(1+cotanx)) ⇒I=∫ (dx/(1+((cosx)/(sinx)))) =∫((sinx)/(sinx +cosx))dx changement tan((x/2))=t give I=∫ (((2t)/(1+t^2 ))/(((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))))((2dt)/(1+t^2 )) I=∫ ((2t)/((1+t^2 )^2 {((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))}))dt =∫ ((2t)/((1+t^2 )(−t^2 +2t+1)))dt =−2∫(t/((t^2 +1)(t^2 −2t−1)))dt let decompose F(t)=(t/((t^2 +1)(t^2 −2t−1))) t^2 −2t−1=0→Δ^′ =1+1=2 ⇒t_1 =1+(√2) and t_2 =1−(√2) F(t)=(t/((t−t_1 )(t−t_2 )(t^2 +1))) =(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a=lim_(t→t_1 ) (t−t_1 )F(t)=(t_1 /((t_1 −t_2 )(t_1 ^2 +1))) =((1+(√2))/(2(√2)(3+2(√2)+1))) =((2+(√2))/(4(4+2(√2)))) =((2+(√2))/(8(2+(√2)))) =(1/8) b=(t_2 /((t_2 −t_1 )(t_2 ^2 +1))) =((1−(√2))/((−2(√2))(3−2(√2)+1))) =((−1+(√2))/(2(√2)(4−2(√2)))) =((−(√2)+2)/(8(2−(√2)))) =(1/8) lim_(t→+∞) tF(t)=0=a+b+c ⇒c=−a−b =−(1/4) ⇒ F(t)=(1/(8(t−t_1 ))) +(1/(8(t−t_2 ))) +((−(1/4)t+d)/(t^2 +1)) F(0)=0 =−(1/(8t_1 ))−(1/(8t_2 )) +d ⇒d=(1/8)((1/t_1 )+(1/t_2 ))=(1/8)(((t_1 +t_2 )/(t_1 t_2 ))) =(1/8)(2/(−1)) =−(1/4) ⇒F(t)=(1/(8(t−t_1 ))) +(1/(8(t−t_2 )))−(1/4) ((t+1)/(t^2 +1)) ⇒ I =−(1/4)∫ (dt/(t−t_1 ))−(1/4)∫ (dt/(t−t_2 )) +(1/2)∫ ((t+1)/(t^2 +1))dt =−(1/4)ln∣t−t_1 ∣−(1/4)ln∣t−t_2 ∣+(1/4)ln(t^2 +1) +(1/2)arctant +c =−(1/4)ln∣tan((x/2))−1−(√2)∣−(1/4)ln∣tan((x/2))−1+(√2)∣ +(1/4)ln(1+tan^2 ((x/2))) +(x/4) +c .$
$l e t I = \int \frac{d x}{1 + c o t a n x} \Rightarrow I = \int \frac{d x}{1 + \frac{c o s x}{s i n x}} = \int \frac{s i n x}{s i n x + c o s x} d x$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e I = \int \frac{\frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $I = \int \frac{2 t}{{(1 + t^{2})}^{2} {\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}}} d t = \int \frac{2 t}{(1 + t^{2}) (- t^{2} + 2 t + 1)} d t$ $= - 2 \int \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)} d t l e t d e c o m p o s e F (t) = \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)}$ $t^{2} - 2 t - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2}$ $F (t) = \frac{t}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c t + d}{t^{2} + 1}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{t_{1}}{(t_{1} - t_{2}) (t_{1}^{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (3 + 2 \sqrt{2} + 1)}$ $= \frac{2 + \sqrt{2}}{4 (4 + 2 \sqrt{2})} = \frac{2 + \sqrt{2}}{8 (2 + \sqrt{2})} = \frac{1}{8}$ $b = \frac{t_{2}}{(t_{2} - t_{1}) (t_{2}^{2} + 1)} = \frac{1 - \sqrt{2}}{(- 2 \sqrt{2}) (3 - 2 \sqrt{2} + 1)} = \frac{- 1 + \sqrt{2}}{2 \sqrt{2} (4 - 2 \sqrt{2})}$ $= \frac{- \sqrt{2} + 2}{8 (2 - \sqrt{2})} = \frac{1}{8}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b + c \Rightarrow c = - a - b = - \frac{1}{4} \Rightarrow$ $F (t) = \frac{1}{8 (t - t_{1})} + \frac{1}{8 (t - t_{2})} + \frac{- \frac{1}{4} t + d}{t^{2} + 1}$ $F (0) = 0 = - \frac{1}{8 t_{1}} - \frac{1}{8 t_{2}} + d \Rightarrow d = \frac{1}{8} (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{1}{8} (\frac{t_{1} + t_{2}}{t_{1} t_{2}})$ $= \frac{1}{8} \frac{2}{- 1} = - \frac{1}{4} \Rightarrow F (t) = \frac{1}{8 (t - t_{1})} + \frac{1}{8 (t - t_{2})} - \frac{1}{4} \frac{t + 1}{t^{2} + 1} \Rightarrow$ $I = - \frac{1}{4} \int \frac{d t}{t - t_{1}} - \frac{1}{4} \int \frac{d t}{t - t_{2}} + \frac{1}{2} \int \frac{t + 1}{t^{2} + 1} d t$ $= - \frac{1}{4} l n ∣ t - t_{1} ∣ - \frac{1}{4} l n ∣ t - t_{2} ∣ + \frac{1}{4} l n (t^{2} + 1) + \frac{1}{2} a r c t a n t + c$ $= - \frac{1}{4} l n ∣ t a n (\frac{x}{2}) - 1 - \sqrt{2} ∣ - \frac{1}{4} l n ∣ t a n (\frac{x}{2}) - 1 + \sqrt{2} ∣$ $+ \frac{1}{4} l n (1 + {t a n}^{2} (\frac{x}{2})) + \frac{x}{4} + c .$
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

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