calculate I = ∫_(π/3) ^(π/2) ((cos(2x))/(sin(x)+cosx))dx and J =∫_(π/3) ^(π/2) ((sin(2x))/(sin(x) +cos(x)))dx


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Question Number 42628 by prof Abdo imad last updated on 29/Aug/18

$c a l c u l a t e I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{c o s (2 x)}{s i n (x) + c o s x} d x a n d$ $J = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{s i n (2 x)}{s i n (x) + c o s (x)} d x$
Commented by maxmathsup by imad last updated on 30/Aug/18
$we have I = ∫_(π/3) ^(π/2) ((cos(2x))/((√2)cos(x−(π/4))))dx changement x−(π/4) =t give I = ∫_(π/(12)) ^(π/2) ((cos(2(t+(π/4))))/((√2)cost)) dt = ∫_(π/(12)) ^(π/2) ((−sin(2t))/((√2)cos(t))) dt =−(1/(√2)) ∫_(π/(12)) ^(π/2) ((2sint cost)/(cost))dt =−(√2) ∫_(π/(12)) ^(π/2) sin(t)dt =−(√2)[−cost]_(π/(12)) ^(π/2) =(√2){0−cos((π/(12)))} we have cos^2 ((π/(12)))=((1+cos((π/6)))/2) =((1+((√3)/2))/2) =((2+(√3))/4) ⇒cos((π/(12)))=((√(2+(√3)))/2) ⇒ I =−(√2)(√(2+(√3)))=−(√(4+2(√3))).$
$w e h a v e I = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{c o s (2 x)}{\sqrt{2} c o s (x - \frac{π}{4})} d x c h a n g e m e n t x - \frac{π}{4} = t g i v e$ $I = \int_{\frac{π}{12}}^{\frac{π}{2}} \frac{c o s (2 (t + \frac{π}{4}))}{\sqrt{2} c o s t} d t = \int_{\frac{π}{12}}^{\frac{π}{2}} \frac{- s i n (2 t)}{\sqrt{2} c o s (t)} d t = - \frac{1}{\sqrt{2}} \int_{\frac{π}{12}}^{\frac{π}{2}} \frac{2 s i n t c o s t}{c o s t} d t$ $= - \sqrt{2} \int_{\frac{π}{12}}^{\frac{π}{2}} s i n (t) d t = - \sqrt{2} {[- c o s t]}_{\frac{π}{12}}^{\frac{π}{2}} = \sqrt{2} {0 - c o s (\frac{π}{12})}$ $w e h a v e {c o s}^{2} (\frac{π}{12}) = \frac{1 + c o s (\frac{π}{6})}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} \Rightarrow c o s (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2} \Rightarrow$ $I = - \sqrt{2} \sqrt{2 + \sqrt{3}} = - \sqrt{4 + 2 \sqrt{3}} .$
Commented by maxmathsup by imad last updated on 30/Aug/18
$we have J = ∫_(π/3) ^(π/2) ((sin(2x))/((√2)sin(x+(π/4))))dx =_(x+(π/4)=t) ∫_((7π)/(12)) ^((3π)/4) ((sin(2(t−(π/4))))/((√2)sint))dt = ∫_((7π)/(12)) ^((3π)/4) ((−cos(2t))/((√2)sint))dt = − ∫_((7π)/(12)) ^((3π)/4) ((1−2sin^2 t)/((√2)sint))dt = ∫_((7π)/(12)) ^((3π)/4) (√2)sint dt−(1/(√2)) ∫_((7π)/(12)) ^((3π)/4) (dt/(sint)) but ∫_((7π)/(12)) ^((3π)/4) (√2)sint dt =(√2)[−cost]_((7π)/(12)) ^((3π)/4) =(√2){ cos(((7π)/(12)))−cos(((3π)/4))} =(√2){ −sin((π/(12))) +cos((π/4))}=(√2){ (1/(√2)) −((√(2−(√3)))/2)} also ∫_((7π)/(12)) ^((3π)/4) (dt/(sint)) =_(tan((t/2)) =u) ∫_(tan(((7π)/(24)))) ^(tan(((3π)/8))) (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) = [ln∣u∣]_(tan(((7π)/(24)))) ^(tan(((3π)/8))) =ln∣tan(((3π)/8))∣−ln∣tan(((7π)/(24)))∣ ⇒ J =(√2){(1/(√2)) −((√(2−(√3)))/2)} −(1/(√2)){ ln∣tan(((3π)/8))∣ −ln∣tan(((7π)/(24)))∣ } .$
$w e h a v e J = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{s i n (2 x)}{\sqrt{2} s i n (x + \frac{π}{4})} d x =_{x + \frac{π}{4} = t} \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \frac{s i n (2 (t - \frac{π}{4}))}{\sqrt{2} s i n t} d t$ $= \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \frac{- c o s (2 t)}{\sqrt{2} s i n t} d t = - \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \frac{1 - 2 {s i n}^{2} t}{\sqrt{2} s i n t} d t = \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \sqrt{2} s i n t d t - \frac{1}{\sqrt{2}} \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \frac{d t}{s i n t}$ $b u t \int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \sqrt{2} s i n t d t = \sqrt{2} {[- c o s t]}_{\frac{7 π}{12}}^{\frac{3 π}{4}} = \sqrt{2} {c o s (\frac{7 π}{12}) - c o s (\frac{3 π}{4})}$ $= \sqrt{2} {- s i n (\frac{π}{12}) + c o s (\frac{π}{4})} = \sqrt{2} {\frac{1}{\sqrt{2}} - \frac{\sqrt{2 - \sqrt{3}}}{2}} a l s o$ $\int_{\frac{7 π}{12}}^{\frac{3 π}{4}} \frac{d t}{s i n t} =_{t a n (\frac{t}{2}) = u} \int_{t a n (\frac{7 π}{24})}^{t a n (\frac{3 π}{8})} \frac{1}{\frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = {[l n ∣ u ∣]}_{t a n (\frac{7 π}{24})}^{t a n (\frac{3 π}{8})}$ $= l n ∣ t a n (\frac{3 π}{8}) ∣ - l n ∣ t a n (\frac{7 π}{24}) ∣ \Rightarrow$ $J = \sqrt{2} {\frac{1}{\sqrt{2}} - \frac{\sqrt{2 - \sqrt{3}}}{2}} - \frac{1}{\sqrt{2}} {l n ∣ t a n (\frac{3 π}{8}) ∣ - l n ∣ t a n (\frac{7 π}{24}) ∣} .$
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