Question Number 42628 by prof Abdo imad last updated on 29/Aug/18
$${calculate}\:\:{I}\:\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{{sin}\left({x}\right)+{cosx}}{dx}\:{and} \\ $$$${J}\:=\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left({x}\right)\:+{cos}\left({x}\right)}{dx} \\ $$
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))).](https://www.tinkutara.com/question/Q42652.png)
$${we}\:{have}\:{I}\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{2}}{cos}\left({x}−\frac{\pi}{\mathrm{4}}\right)}{dx}\:\:\:{changement}\:{x}−\frac{\pi}{\mathrm{4}}\:={t}\:{give}\:\:\:\:\:\: \\ $$$${I}\:=\:\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cos}\left(\mathrm{2}\left({t}+\frac{\pi}{\mathrm{4}}\right)\right)}{\:\sqrt{\mathrm{2}}{cost}}\:{dt}\:\:=\:\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{−{sin}\left(\mathrm{2}{t}\right)}{\:\sqrt{\mathrm{2}}{cos}\left({t}\right)}\:{dt}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}{sint}\:{cost}}{{cost}}{dt} \\ $$$$=−\sqrt{\mathrm{2}}\:\:\:\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}\left({t}\right){dt}\:=−\sqrt{\mathrm{2}}\left[−{cost}\right]_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{2}}} \:\:=\sqrt{\mathrm{2}}\left\{\mathrm{0}−{cos}\left(\frac{\pi}{\mathrm{12}}\right)\right\} \\ $$$${we}\:{have}\:{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)=\frac{\mathrm{1}+{cos}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}}\:=\frac{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}}\:=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rightarrow{cos}\left(\frac{\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=−\sqrt{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}=−\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}. \\ $$
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)))∣ } .](https://www.tinkutara.com/question/Q42653.png)
$${we}\:{have}\:{J}\:=\:\int_{\frac{\pi}{\mathrm{3}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}{dx}\:\:=_{{x}+\frac{\pi}{\mathrm{4}}={t}} \:\:\:\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\frac{{sin}\left(\mathrm{2}\left({t}−\frac{\pi}{\mathrm{4}}\right)\right)}{\:\sqrt{\mathrm{2}}{sint}}{dt}\:\:\:\:\:\:\:\: \\ $$$$=\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\:\frac{−{cos}\left(\mathrm{2}{t}\right)}{\:\sqrt{\mathrm{2}}{sint}}{dt}\:=\:−\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\:\frac{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {t}}{\:\sqrt{\mathrm{2}}{sint}}{dt}\:=\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\sqrt{\mathrm{2}}{sint}\:{dt}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\frac{{dt}}{{sint}} \\ $$$$\:{but}\:\:\:\:\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\sqrt{\mathrm{2}}{sint}\:{dt}\:=\sqrt{\mathrm{2}}\left[−{cost}\right]_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:=\sqrt{\mathrm{2}}\left\{\:{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{12}}\right)−{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\sqrt{\mathrm{2}}\left\{\:−{sin}\left(\frac{\pi}{\mathrm{12}}\right)\:+{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right\}=\sqrt{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\right\}\:{also} \\ $$$$\int_{\frac{\mathrm{7}\pi}{\mathrm{12}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\:\:\frac{{dt}}{{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}} \:\:\:\int_{{tan}\left(\frac{\mathrm{7}\pi}{\mathrm{24}}\right)} ^{{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\left[{ln}\mid{u}\mid\right]_{{tan}\left(\frac{\mathrm{7}\pi}{\mathrm{24}}\right)} ^{{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)} \\ $$$$={ln}\mid{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\mid−{ln}\mid{tan}\left(\frac{\mathrm{7}\pi}{\mathrm{24}}\right)\mid\:\:\Rightarrow \\ $$$${J}\:=\sqrt{\mathrm{2}}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:−\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\right\}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\:{ln}\mid{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\mid\:−{ln}\mid{tan}\left(\frac{\mathrm{7}\pi}{\mathrm{24}}\right)\mid\:\right\}\:. \\ $$$$ \\ $$