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Question Number 63410 by aliesam last updated on 03/Jul/19

Commented by mathmax by abdo last updated on 03/Jul/19
$2) let I =∫ ((xdx)/(1+sinx)) changement tan((x/2))=t give I = ∫ ((2arctan(t))/((1+((2t)/(1+t^2 )))(1+t^2 )))dt = 2∫ ((arctan(t))/(1+t^2 +2t)) dt =2 ∫ ((arctan(t))/((t+1)^2 )) dt =_(by pwrts) 2{ −((arcant)/(t+1)) −∫ −(1/(t+1)) (dt/(1+t^2 ))} −((2 arctan(t))/(t+1)) + 2 ∫ (dt/((t+1)(t^2 +1))) let decompose F(t) =(1/((t+1)(t^2 +1))) ⇒F(t) =(a/(t+1)) +((bt +c)/(t^2 +1)) a =lim_(t→−1) (t+1)F(t) =(1/2) lim_(t→+∞) t F(t) =0 =a+b ⇒b=−(1/2) ⇒F(t)=(1/(2(t+1))) −(1/2) ((t −2c)/(t^2 +1)) F(0) =1 =(1/2) +c ⇒c=(1/2) ⇒F(t)=(1/(2(t+1))) −(1/2) ((t−1)/(t^2 +1)) ⇒ ∫ (dt/((t+1)(t^2 +1))) = ∫ (dt/(2(t+1))) −(1/4) ∫ ((2t)/(t^2 +1)) +(1/2) ∫ (dt/(t^2 +1)) =(1/2)ln∣t+1∣−(1/4)ln(t^2 +1)+(1/2) arctan(t)+c =(1/2)ln∣1+tan((x/2))∣−(1/4)ln(1+tan^2 ((x/2))) +x +c ⇒ I =−(x/(1+tan((x/2)))) +ln∣1+tan((x/2))∣−(1/2)ln(1+tan^2 ((x/2))) +x +c .$
$2) l e t I = \int \frac{x d x}{1 + s i n x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{2 a r c t a n (t)}{(1 + \frac{2 t}{1 + t^{2}}) (1 + t^{2})} d t = 2 \int \frac{a r c t a n (t)}{1 + t^{2} + 2 t} d t = 2 \int \frac{a r c t a n (t)}{{(t + 1)}^{2}} d t$ $=_{b y p w r t s} 2 {- \frac{a r c a n t}{t + 1} - \int - \frac{1}{t + 1} \frac{d t}{1 + t^{2}}}$ $- \frac{2 a r c t a n (t)}{t + 1} + 2 \int \frac{d t}{(t + 1) (t^{2} + 1)} l e t d e c o m p o s e$ $F (t) = \frac{1}{(t + 1) (t^{2} + 1)} \Rightarrow F (t) = \frac{a}{t + 1} + \frac{b t + c}{t^{2} + 1}$ $a = {l i m}_{t \to - 1} (t + 1) F (t) = \frac{1}{2}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - \frac{1}{2} \Rightarrow F (t) = \frac{1}{2 (t + 1)} - \frac{1}{2} \frac{t - 2 c}{t^{2} + 1}$ $F (0) = 1 = \frac{1}{2} + c \Rightarrow c = \frac{1}{2} \Rightarrow F (t) = \frac{1}{2 (t + 1)} - \frac{1}{2} \frac{t - 1}{t^{2} + 1} \Rightarrow$ $\int \frac{d t}{(t + 1) (t^{2} + 1)} = \int \frac{d t}{2 (t + 1)} - \frac{1}{4} \int \frac{2 t}{t^{2} + 1} + \frac{1}{2} \int \frac{d t}{t^{2} + 1}$ $= \frac{1}{2} l n ∣ t + 1 ∣ - \frac{1}{4} l n (t^{2} + 1) + \frac{1}{2} a r c t a n (t) + c$ $= \frac{1}{2} l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{4} l n (1 + {t a n}^{2} (\frac{x}{2})) + x + c \Rightarrow$ $I = - \frac{x}{1 + t a n (\frac{x}{2})} + l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{2} l n (1 + {t a n}^{2} (\frac{x}{2})) + x + c .$
Commented by aliesam last updated on 04/Jul/19

$p e r f e c t$
	Answered by MJS last updated on 04/Jul/19

	$2)$ $\int \frac{x}{1 + \sin x} d x$ $by parts$ $u^{'} = \frac{1}{1 + \sin x} \to u = \int \frac{d x}{1 + \sin x}$ $v = x \to v^{'} = 1$ $\int u^{'} v = u v - \int {u v}^{'}$ $\int \frac{x}{1 + \sin x} d x = x \int \frac{d x}{1 + \sin x} - \int (\int \frac{d x}{1 + \sin x}) d x$ $\int \frac{d x}{1 + \sin x} =$ $[t = \tan \frac{x}{2} \to d x = {2 \cos}^{2} \frac{x}{2} d t]$ $= 2 \int \frac{d t}{{(t + 1)}^{2}} = - \frac{2}{t + 1} = - 2 \frac{\cos \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} =$ $= \tan x - \frac{1}{\cos x} - 1$ $\int (\int \frac{d x}{1 + \sin x}) d x = \int \tan x d x - \int \frac{d x}{\cos x} - \int d x =$ $= - \ln \cos x - \ln \frac{\cos x}{1 - \sin x} - x = - x + \ln \frac{1}{1 + \sin x}$ $\Rightarrow$ $\int \frac{x}{1 + \sin x} d x = x \tan x - \frac{x}{\cos x} - x + x - \ln \frac{1}{1 + \sin x} =$ $= x \tan x - \frac{x}{\cos x} - \ln \frac{1}{1 + \sin x} + C$
	Answered by MJS last updated on 04/Jul/19

	$1)$ $\int \frac{d x}{\sqrt{1 + \tan x}} =$ $[t = \sqrt{1 + \tan x} \to d x = \frac{2 t}{t^{4} - 2 t^{2} + 2} d t]$ $= 2 \int \frac{d t}{t^{4} - 2 t^{2} + 2} = 2 \int \frac{d t}{(t^{2} - \sqrt{2 + 2 \sqrt{2}} t + \sqrt{2}) (t^{2} + \sqrt{2 + 2 \sqrt{2}} t + \sqrt{2})} =$ $\sqrt{2 + 2 \sqrt{2}} = a$ $\sqrt{2} = b$ $= 2 \int \frac{d t}{(t^{2} - a t + b) (t^{2} + a t + b)} =$ $= - \frac{1}{a b} \int \frac{t - a}{t^{2} - a t + b} d t + \frac{1}{a b} \int \frac{t + a}{t^{2} + a t + b} d t =$ $= - \frac{1}{2 a b} \int \frac{2 t - a}{t^{2} - a t + b} d t + \frac{1}{2 b} \int \frac{d t}{t^{2} - a t + b} + \frac{1}{2 a b} \int \frac{2 t + a}{t^{2} + a t + b} d t + \frac{1}{2 b} \int \frac{d t}{t^{2} + a t + b} =$ $= - \frac{1}{2 a b} \ln (t^{2} - a t + b) + \frac{1}{b \sqrt{4 b - a^{2}}} \arctan \frac{2 t - a}{\sqrt{4 b - a^{2}}} + \frac{1}{2 a b} \ln (t^{2} + a t + b) + \frac{1}{b \sqrt{4 b - a^{2}}} \arctan \frac{2 t + a}{\sqrt{4 b - a^{2}}} =$ $= \frac{1}{2 a b} \ln \frac{t^{2} + a t + b}{t^{2} - a t + b} + \frac{1}{b \sqrt{4 b - a^{2}}} (\arctan \frac{2 t - a}{\sqrt{4 b - a^{2}}} + \arctan \frac{2 t + a}{\sqrt{4 b - a^{2}}})$ $. . .$
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