Question Number 12201 by Nayon last updated on 16/Apr/17
$$\int\frac{\mathrm{1}}{{sin}\left({x}\right)}{dx} \\ $$
Answered by ajfour last updated on 16/Apr/17
$${I}=\int\:\frac{\left(\mathrm{cosec}\:{x}\right)\left(\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}\right)}{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}}{dx} \\ $$$${let}\:\:\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}\:=\:{t} \\ $$$$\frac{{dt}}{{dx}}=\:−\left(\mathrm{cosec}\:{x}\right)\mathrm{cot}\:{x}−\mathrm{cosec}\:^{\mathrm{2}} {x} \\ $$$$\left(\mathrm{cosec}\:{x}\right)\left(\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}\right){dx}=−{dt} \\ $$$${I}\:=\:−\int\frac{{dt}}{{t}}\:=\:−\mathrm{ln}\:\mid{t}\mid+{C} \\ $$$$\int\frac{{dx}}{\mathrm{sin}\:{x}}\:=\:−\mathrm{ln}\:\mid\mathrm{cosec}\:{x}\:+\:\mathrm{cot}\:{x}\mid+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{ln}\:\mid\mathrm{cosec}\:{x}−\mathrm{cot}\:{x}\mid+{C}\:. \\ $$
Answered by mrW1 last updated on 16/Apr/17
![∫(1/(sin x)) dx =∫((sin x)/(sin^2 x)) dx =∫((dcos x)/(cos^2 x−1)) =∫(dt/(t^2 −1)) (t=cos x) =(1/2)[∫(dt/(t−1))−∫(dt/(t+1))] =(1/2)ln (((1−t)/(1+t)))+C =(1/2)ln (((1−cos x)/(1+cos x)))+C =(1/2)ln (((1−1+2sin^2 (x/2))/(1+2cos^2 (x/2)−1)))+C =(1/2)ln (((sin^2 (x/2))/(cos^2 (x/2))))+C =ln ∣tan (x/2)∣+C](https://www.tinkutara.com/question/Q12205.png)
$$\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:{dx} \\ $$$$=\int\frac{\mathrm{sin}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}\:{dx} \\ $$$$=\int\frac{{d}\mathrm{cos}\:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{1}} \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}\:\:\:\:\:\:\:\left({t}=\mathrm{cos}\:{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{{dt}}{{t}−\mathrm{1}}−\int\frac{{dt}}{{t}+\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{1}−\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}{\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}−\mathrm{1}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}{\mathrm{cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}\right)+{C} \\ $$$$=\mathrm{ln}\:\mid\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\mid+{C} \\ $$
Answered by mrW1 last updated on 16/Apr/17
![∫(1/(sin x)) dx =∫(1/(2sin (x/2)cos (x/2))) dx =∫(1/(sin (x/2)cos (x/2))) d(x/2) =∫(1/(sin ucos u)) du (u=(x/2)) =∫((cos u)/(sin ucos^2 u)) du =∫((cos u)/(sin u(1−sin^2 u))) du =∫(1/(sin u(1−sin^2 u))) dsin u =∫(1/(v(1−v^2 ))) dv (v=sin u) =(1/2)∫(1/(v^2 (1−v^2 ))) dv^2 =(1/2)∫(1/(w(1−w))) dw (w=v^2 =sin^2 u) =(1/2)[∫(1/w)dw+∫(1/(1−w))dw] =(1/2)ln (w/(1−w))+C =(1/2)ln ((sin^2 u)/(1−sin^2 u))+C =(1/2)ln ((sin^2 u)/(cos^2 u))+C =ln ∣tan u∣+C =ln ∣tan (x/2)∣+C](https://www.tinkutara.com/question/Q12206.png)
$$\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\:{d}\frac{{x}}{\mathrm{2}} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{sin}\:{u}\mathrm{cos}\:{u}}\:{du}\:\:\:\:\:\:\:\left({u}=\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\int\frac{\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}\mathrm{cos}^{\mathrm{2}} \:{u}}\:{du} \\ $$$$=\int\frac{\mathrm{cos}\:{u}}{\mathrm{sin}\:{u}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{u}\right)}\:{du} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{sin}\:{u}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{u}\right)}\:{d}\mathrm{sin}\:{u} \\ $$$$=\int\frac{\mathrm{1}}{{v}\left(\mathrm{1}−{v}^{\mathrm{2}} \right)}\:{dv}\:\:\:\:\:\:\:\:\:\:\:\:\left({v}=\mathrm{sin}\:{u}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{v}^{\mathrm{2}} \left(\mathrm{1}−{v}^{\mathrm{2}} \right)}\:{dv}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{w}\left(\mathrm{1}−{w}\right)}\:{dw}\:\:\:\:\:\:\:\:\:\left({w}={v}^{\mathrm{2}} =\mathrm{sin}^{\mathrm{2}} \:{u}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{\mathrm{1}}{{w}}{dw}+\int\frac{\mathrm{1}}{\mathrm{1}−{w}}{dw}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{w}}{\mathrm{1}−{w}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{sin}^{\mathrm{2}} \:{u}}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{u}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{sin}^{\mathrm{2}} \:{u}}{\mathrm{cos}^{\mathrm{2}} \:\:{u}}+{C} \\ $$$$=\mathrm{ln}\:\mid\mathrm{tan}\:{u}\mid+{C} \\ $$$$=\mathrm{ln}\:\mid\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\mid+{C} \\ $$