find the value of ∫_0 ^(π/4) ((cosx)/(sinx +tanx))dx


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Question Number 36057 by abdo mathsup 649 cc last updated on 28/May/18

$f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} \frac{c o s x}{s i n x + t a n x} d x$
Commented by abdo mathsup 649 cc last updated on 30/May/18
$let I = ∫_0 ^(π/4) ((cosx)/(sinx +tanx))dx I = ∫_0 ^(π/4) ((cosx)/(sinx +((sinx)/(cosx))))dx = ∫_0 ^(π/4) ((cos^2 x)/(sinx.cosx +sinx))dx changement tan((x/2)) =t give I = ∫_0 ^((√2) −1) (((((1−t^2 )/(1+t^2 )))^2 )/(((2t (1−t^(2)) )/((1+t^2 )^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 2∫_0 ^((√2) −1) (((1−t^2 )^2 )/(2t(1−t^2 ) +2t(1+t^2 ))) (dt/(1+t^2 )) =2 ∫_0 ^((√2) −1) (((1−t^2 )^2 )/((1+t^2 )( 2t −2t^3 +2t +2t^3 )))dt = (1/2) ∫_0 ^((√2) −1) (((1−t^2 )^2 )/((1+t^2 )t)) dt ⇒ 2I = ∫_0 ^((√2)−1) ((t^4 −2t^2 +1)/(t^3 +t))dt =∫_0 ^((√2)−1) ((t(t^3 +t)−t^2 −2t^2 +1)/(t^(3 ) +t))dt = ∫_0 ^((√2)−1) t dt +∫_0 ^((√2) −1) ((−3t^2 +1)/(t^3 +t))dt but ∫_0 ^((√2) −1) tdt =[(t^2 /2)]_0 ^((√2)−1) =((((√2)−1)^2 )/2) let decompose F(t) = ((−3t^2 +1)/(t^3 +t)) F(t) = ((−3t^2 +1)/(t^3 +t)) = (a/t) +((bt +c)/(t^2 +1)) a=lim_(t→0) tF(t) = 1 =a lim_(t→+∞) t F(t) = −3 =a +b ⇒b=−3−a=−4 F(t) = (1/t) +((−4t +c)/(t^(2 ) +1)) F(1) =−1 = 1 +((−4+c)/2) =−1 +(c/2) ⇒c=0 ⇒ F(t)= (1/t) −((4t)/(t^2 +1)) ⇒ ∫_0 ^((√2) −1) ((−3t^2 +1)/(t^3 +t))dt = ∫_0 ^((√2)−1) { (1/t) −((4t)/(t^2 +1))}dt =[ln∣t∣ −2 ln(t^2 +1)]_0 ^((√2) −1) = [ln( (t/((t^2 +1)^2 )))]_0 ^((√2)−1) = ln((((√2) −1)/({((√2) −1)^2 +1}^( ))) −ln(0^+ ) =+∞ so this integral is divergent !$
$l e t I = \int_{0}^{\frac{π}{4}} \frac{c o s x}{s i n x + t a n x} d x$ $I = \int_{0}^{\frac{π}{4}} \frac{c o s x}{s i n x + \frac{s i n x}{c o s x}} d x = \int_{0}^{\frac{π}{4}} \frac{{c o s}^{2} x}{s i n x . c o s x + s i n x} d x$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int_{0}^{\sqrt{2} - 1} \frac{{(\frac{1 - t^{2}}{1 + t^{2}})}^{2}}{\frac{2 t (1 - t^{2)}}{{(1 + t^{2})}^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{{(1 - t^{2})}^{2}}{2 t (1 - t^{2}) + 2 t (1 + t^{2})} \frac{d t}{1 + t^{2}}$ $= 2 \int_{0}^{\sqrt{2} - 1} \frac{{(1 - t^{2})}^{2}}{(1 + t^{2}) (2 t - 2 t^{3} + 2 t + 2 t^{3})} d t$ $= \frac{1}{2} \int_{0}^{\sqrt{2} - 1} \frac{{(1 - t^{2})}^{2}}{(1 + t^{2}) t} d t \Rightarrow$ $2 I = \int_{0}^{\sqrt{2} - 1} \frac{t^{4} - 2 t^{2} + 1}{t^{3} + t} d t$ $= \int_{0}^{\sqrt{2} - 1} \frac{t (t^{3} + t) - t^{2} - 2 t^{2} + 1}{t^{3} + t} d t$ $= \int_{0}^{\sqrt{2} - 1} t d t + \int_{0}^{\sqrt{2} - 1} \frac{- 3 t^{2} + 1}{t^{3} + t} d t b u t$ $\int_{0}^{\sqrt{2} - 1} t d t = {[\frac{t^{2}}{2}]}_{0}^{\sqrt{2} - 1} = \frac{{(\sqrt{2} - 1)}^{2}}{2} l e t d e c o m p o s e$ $F (t) = \frac{- 3 t^{2} + 1}{t^{3} + t}$ $F (t) = \frac{- 3 t^{2} + 1}{t^{3} + t} = \frac{a}{t} + \frac{b t + c}{t^{2} + 1}$ $a = {l i m}_{t \to 0} t F (t) = 1 = a$ ${l i m}_{t \to + \infty} t F (t) = - 3 = a + b \Rightarrow b = - 3 - a = - 4$ $F (t) = \frac{1}{t} + \frac{- 4 t + c}{t^{2} + 1}$ $F (1) = - 1 = 1 + \frac{- 4 + c}{2} = - 1 + \frac{c}{2} \Rightarrow c = 0 \Rightarrow$ $F (t) = \frac{1}{t} - \frac{4 t}{t^{2} + 1} \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} \frac{- 3 t^{2} + 1}{t^{3} + t} dt = \int_{0}^{\sqrt{2} - 1} {\frac{1}{t} - \frac{4 t}{t^{2} + 1}} d t$ $= {[l n ∣ t ∣ - 2 l n (t^{2} + 1)]}_{0}^{\sqrt{2} - 1}$ $= {[l n (\frac{t}{{(t^{2} + 1)}^{2}})]}_{0}^{\sqrt{2} - 1} = l n (\frac{\sqrt{2} - 1}{{{(\sqrt{2} - 1)}^{2} + 1}^{(}}) - l n (0^{+})$ $= + \infty s o t h i s i n t e g r a l i s d i v e r g e n t!$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18

	$\int_{0}^{\frac{Π}{4}} \frac{{c o s}^{2} x}{s i n x c o s x + s i n x} d x$ $(- 1) \int_{0}^{\frac{Π}{4}} \frac{1 - {c o s}^{2} x - 1}{s i n x (1 + c o s x)} d x$ $(- 1) \int_{0}^{\frac{Π}{4}} \frac{1 - c o s x}{s i n x} d x + \int_{0}^{\frac{Π}{4}} \frac{s i n x}{{s i n}^{2} x (1 + c o s x)} d x$ $\int_{0}^{\frac{Π}{4}} \frac{c o t x - c o s e c x}{} - \int_{1}^{\frac{1}{\sqrt{2}}} \frac{d t}{(1 - t^{2}) (1 + t)}$ $∣ l n s i n x - l n t a n \frac{x}{2} ∣_{0}^{\frac{Π}{4}} + I_{2}$ $I_{2} = \int_{1}^{\frac{1}{\sqrt{2}}} \frac{A}{1 + t} + \frac{B}{1 - t} + \frac{C}{{(1 + t)}^{2}} d t$ $\frac{1}{{(1 + t)}^{2} (1 - t)} = \frac{A}{1 + t} + \frac{B}{1 - t} + \frac{C}{{(1 + t)}^{2}}$ $1 = A (1 + t) (1 - t) + B {(1 + t)}^{2} + C (1 - t)$ $1 = A (1 - t^{2}) + B (1 + 2 t + t^{2}) + C (1 - t)$ $A - {A t}^{2} + B + 2 B t + {B t}^{2} + C - C t = 1$ $A + B + C + t (2 B - C) + t^{2} (B - A) = 1$ $A + B + C = 1$ $2 B - C = 0$ $B - A = 0$ $A = B$ $2 B - C = 0$ $C = 2 B$ $A + B + C = 1$ $B + B + 2 B = 1$ $B = \frac{1}{4} A = \frac{1}{4} C = \frac{1}{2}$ $\int_{1}^{\frac{1}{\sqrt{2}}} \frac{\frac{1}{4}}{1 + t} d t + \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\frac{1}{4}}{1 - t} d t + \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\frac{1}{2}}{{(1 + t)}^{2}} d t$ $∣ \frac{1}{4} l n (1 + t) - \frac{1}{4} l n (1 - t) - \frac{1}{2} \times \frac{1}{1 + t} ∣_{1}^{\frac{1}{\sqrt{2}}}$
	Answered by sma3l2996 last updated on 28/May/18

	$A = \int_{0}^{π / 4} \frac{c o s x}{s i n x + t a n x} d x = \int_{0}^{π / 4} \frac{{c o s}^{2} x}{c o s x s i n x + s i n x} d x$ $= \int_{0}^{π / 4} \frac{1 - {s i n}^{2} x}{s i n x (c o s x + 1)} d x = \int_{0}^{π / 4} \frac{d x}{s i n x (c o s x + 1)} - \int_{0}^{π / 4} \frac{s i n x}{1 + c o s x} d x$ $= \int_{0}^{π / 4} \frac{d x}{s i n x (c o s x + 1)} + {[l n ∣ 1 + c o s x ∣]}_{0}^{π / 4}$ $l e t u = t a n x / 2 \Rightarrow d x = \frac{2 d u}{1 + u^{2}}$ $s i n x = 2 c o s (x / 2) s i n (x / 2) = \frac{2 u}{1 + u^{2}}$ $1 + c o s x = 2 {c o s}^{2} (x / 2) = \frac{2}{1 + u^{2}}$ $A = l n (\frac{\sqrt{2} + 1}{2}) - l n 2 + \int_{0}^{{t a n}^{- 1} (π / 8)} \frac{1}{\frac{2 u}{1 + u^{2}} \times \frac{2}{1 + u^{2}}} \times \frac{2 d u}{1 + u^{2}}$ $A = l n (\sqrt{2} + 2) - 2 l n 2 + \frac{1}{2} \int_{0}^{\sqrt{2} - 1} \frac{1 + u^{2}}{u} d u$ $= l n (\sqrt{2} + 2) - 2 l n 2 + \frac{1}{2} {[l n u + \frac{1}{2} u^{2}]}_{0}^{\sqrt{2} - 1}$ $A = - \infty b e c a u s e (\underset{u \to 0}{l i m l n u} = - \infty)$
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