Question Number 43158 by MASANJA J last updated on 07/Sep/18
$$\int{cosecxdx} \\ $$
Commented by maxmathsup by imad last updated on 07/Sep/18
$${let}\:{I}\:=\:\int\:\:\:\frac{{dx}}{{sin}\left({x}\right)}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\:\:\frac{{dt}}{{t}}\:={ln}\mid{t}\mid\:+{c}\:={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+{c}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18
![∫(dx/(sinx)) ∫((sinx)/(sin^2 x))dx ∫((sinx)/(1−cos^2 x))dx t=cosx dt=−sinx dx ∫((−dt)/(1−t^2 )) ∫(dt/((t+1)(t−1))) (1/2)∫(((t+1)−(t−1))/((t+1)(t−1)))dt (1/2)[∫(dt/(t−1))−∫(dt/(t+1))] (1/2)[ln(t−1)−ln(t+1)] (1/2)ln∣((t−1)/(t+1))∣+c =(1/2)ln∣((cosx−1)/(cosx+1))∣+c =(1/2)ln∣((1−2sin^2 (x/2)−1)/(2cos^2 (x/2)))∣+c =(1/2)ln∣tan^2 (x/2)∣+c =ln(tan(x/2))+c recheck (d/dx){lntan(x/2)}=(1/(tan(x/2)))×sec^2 (x/2)×(1/2) =(1/(2×((sin(x/2))/(cos(x/2)))×cos^2 (x/2))) =(1/(2sin(x/2)×cos(x/2)))=(1/(sinx))=cosecx](https://www.tinkutara.com/question/Q43169.png)
$$\int\frac{{dx}}{{sinx}} \\ $$$$\int\frac{{sinx}}{{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{{sinx}}{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${t}={cosx}\:\:{dt}=−{sinx}\:{dx} \\ $$$$\int\frac{−{dt}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({t}+\mathrm{1}\right)−\left({t}−\mathrm{1}\right)}{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{{dt}}{{t}−\mathrm{1}}−\int\frac{{dt}}{{t}+\mathrm{1}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({t}−\mathrm{1}\right)−{ln}\left({t}+\mathrm{1}\right)\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{cosx}−\mathrm{1}}{{cosx}+\mathrm{1}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\mid+{c} \\ $$$$={ln}\left({tan}\frac{{x}}{\mathrm{2}}\right)+{c} \\ $$$${recheck} \\ $$$$\frac{{d}}{{dx}}\left\{{lntan}\frac{{x}}{\mathrm{2}}\right\}=\frac{\mathrm{1}}{{tan}\frac{{x}}{\mathrm{2}}}×{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}×\frac{{sin}\frac{{x}}{\mathrm{2}}}{{cos}\frac{{x}}{\mathrm{2}}}×{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}×{cos}\frac{{x}}{\mathrm{2}}}=\frac{\mathrm{1}}{{sinx}}={cosecx} \\ $$
Answered by $@ty@m last updated on 08/Sep/18
$$\int\mathrm{cosec}\:{x}×\frac{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}}{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}}{dx} \\ $$$$=−\int\frac{−\mathrm{cosec}\:^{\mathrm{2}} {x}−\mathrm{cosec}\:{x}.\mathrm{cot}{x}}{\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}}{dx} \\ $$$$=−\mathrm{ln}\:\left(\mathrm{cosec}\:{x}+\mathrm{cot}\:{x}\right)+{C} \\ $$