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Question Number 89636 by jagoll last updated on 18/Apr/20
∫ ((cos x+sin x)/(sin 2x)) dx
$$\int\:\frac{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{2x}}\:\mathrm{dx}\: \\ $$
Commented by M±th+et£s last updated on 18/Apr/20
A=∫(1/(cos(x)))+tan(x)−tan(x) dx ∫((1+sin(x))/(cos(x)))dx−∫((sin(x))/(cos(x)))dx ∫((cos(x))/(1−sin(x)))dx−∫((sin(x))/(cos(x)))dx −ln∣1−sin(x)∣+ln∣cos(x)∣+c so∫((cos(x)+sin(x))/(sin(2x)))dx=(1/2)(I+A)+c
$${A}=\int\frac{\mathrm{1}}{{cos}\left({x}\right)}+{tan}\left({x}\right)−{tan}\left({x}\right)\:{dx} \\ $$$$\int\frac{\mathrm{1}+{sin}\left({x}\right)}{{cos}\left({x}\right)}{dx}−\int\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}{dx} \\ $$$$\int\frac{{cos}\left({x}\right)}{\mathrm{1}−{sin}\left({x}\right)}{dx}−\int\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}{dx} \\ $$$$−{ln}\mid\mathrm{1}−{sin}\left({x}\right)\mid+{ln}\mid{cos}\left({x}\right)\mid+{c} \\ $$$$ \\ $$$${so}\int\frac{{cos}\left({x}\right)+{sin}\left({x}\right)}{{sin}\left(\mathrm{2}{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({I}+{A}\right)+{c} \\ $$$$ \\ $$$$ \\ $$
Commented by jagoll last updated on 18/Apr/20
thank you all
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all} \\ $$
Answered by M±th+et£s last updated on 18/Apr/20
(1/2)∫(1/(sin(x)))dx +(1/2)∫(1/(cos(x)))dx I=∫(1/(sin(x)))+cot(x)−cot(x) dx =∫((1+cos(x))/(sin(x)))dx − ∫((cos(x))/(sin(x)))dx =∫((sin(x))/(1−cos(x)))dx −∫((cos(x))/(sin(x)))dx ln∣1−cos(x)∣−ln∣sin(x)∣+c
$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{sin}\left({x}\right)}{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{cos}\left({x}\right)}{dx} \\ $$$${I}=\int\frac{\mathrm{1}}{{sin}\left({x}\right)}+{cot}\left({x}\right)−{cot}\left({x}\right)\:{dx} \\ $$$$=\int\frac{\mathrm{1}+{cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}\:−\:\int\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$ \\ $$$$=\int\frac{{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}{dx}\:−\int\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$${ln}\mid\mathrm{1}−{cos}\left({x}\right)\mid−{ln}\mid{sin}\left({x}\right)\mid+{c} \\ $$
Answered by john santu last updated on 18/Apr/20
cos x+sin x = (√2)cos (x−(π/4)) [ let u = x−(π/4) ⇒ dx = du ] ∫ (((√2) cos u du)/(sin 2(u+(π/4)))) = ∫ (((√2) cos u du)/(cos 2u)) ∫ (((√2) cos u du)/(1−2sin^2 u )) [ w = (√2) sin u ] ∫ (dw/(1−w^2 )) = (1/2)∫ ((1/(1−w)) +(1/(1+w)) ) dw = (1/2)ln ∣((1+w)/(1−w))∣ + c = (1/2) ln ∣((1+(√2) sin u)/(1−(√2) sin u))∣ + c = (1/2)ln ∣((1+(√2) sin (x−(π/4)))/(1−(√2) sin (x−(π/4))))∣ + c
$$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\:=\:\sqrt{\mathrm{2}}\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\: \\ $$$$\left[\:{let}\:{u}\:=\:{x}−\frac{\pi}{\mathrm{4}}\:\Rightarrow\:{dx}\:=\:{du}\:\right] \\ $$$$\int\:\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:{u}\:{du}}{\mathrm{sin}\:\mathrm{2}\left({u}+\frac{\pi}{\mathrm{4}}\right)}\:=\:\int\:\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:{u}\:{du}}{\mathrm{cos}\:\mathrm{2}{u}} \\ $$$$\int\:\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:{u}\:{du}}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {u}\:}\:\left[\:{w}\:=\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:{u}\:\right]\:\: \\ $$$$\int\:\frac{{dw}}{\mathrm{1}−{w}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\left(\frac{\mathrm{1}}{\mathrm{1}−{w}}\:+\frac{\mathrm{1}}{\mathrm{1}+{w}}\:\right)\:{dw} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{1}+{w}}{\mathrm{1}−{w}}\mid\:+\:{c}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:{u}}{\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:{u}}\mid\:+\:{c}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}−\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{4}}\right)}\mid\:+\:{c}\: \\ $$$$ \\ $$
Answered by petrochengula last updated on 18/Apr/20
=−∫((cosx+sinx)/(−sin2x))dx =−∫((cosx+sinx)/(1−1−sin2x))dx =−∫((cosx+sinx)/(1−sin2x−1))dx =−∫((cosx+sinx)/((sinx−cosx)^2 −1))dx let t=sinx−cosx⇒dt=(cosx+sinx)dx =−∫(1/(t^2 −1))dt =−∫(1/((1+t)(1−t)))dt =−(1/2)∫((1+1+t−t)/((1+t)(1+t)))dt =−(1/2)∫((1+t+1−t)/((1+t)(1−t)))dt =−(1/2)∫(1/(1−t))dt−(1/2)∫(1/(1+t))dt =(1/2)ln∣1−t∣−(1/2)ln∣1+t∣+c =(1/2)ln∣1−sinx+cosx∣−(1/2)ln∣1+sinx−cosx∣+c
$$=−\int\frac{{cosx}+{sinx}}{−{sin}\mathrm{2}{x}}{dx} \\ $$$$=−\int\frac{{cosx}+{sinx}}{\mathrm{1}−\mathrm{1}−{sin}\mathrm{2}{x}}{dx} \\ $$$$=−\int\frac{{cosx}+{sinx}}{\mathrm{1}−{sin}\mathrm{2}{x}−\mathrm{1}}{dx} \\ $$$$=−\int\frac{{cosx}+{sinx}}{\left({sinx}−{cosx}\right)^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$${let}\:{t}={sinx}−{cosx}\Rightarrow{dt}=\left({cosx}+{sinx}\right){dx} \\ $$$$=−\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$=−\int\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\mathrm{1}+{t}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}\right)}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{t}+\mathrm{1}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{1}−{t}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}−{t}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{t}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}−{sinx}+{cosx}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{sinx}−{cosx}\mid+{c} \\ $$

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