calculate-A-dt-sin-t-by-using-u-cos-t-

Question Number 169249 by mathocean1 last updated on 26/Apr/22

$${calculate}\:{A}=\int\frac{{dt}}{{sin}\left({t}\right)}\:{by}\:{using} \\ $$$${u}={cos}\left({t}\right) \\ $$

Commented by infinityaction last updated on 26/Apr/22

A = ∫((sin t)/(sin^2 t))dt A = ∫((sin t)/(1−cos^2 t))dt u = cos t ⇒ −du = sin tdt A = −∫(du/(1−u^2 )) ⇒ −∫(du/((1−u)(1+u))) A = ((−1)/2){∫(du/(1−u))+∫ (du/(1+u))} A = ((−1)/2){−log (1−u)+log (1+u)} +c A = (1/2)log∣ ((1−u)/(1+u)) ∣+ c A = (1/2)log ∣((1−cos t)/(1+cos t))∣ +c A = (1/2)log ∣tan^2 (t/2)∣ + c A = log ∣tan (t/2)∣ + c

$$\:\:\:\:\:\:\:{A}\:=\:\int\frac{\mathrm{sin}\:{t}}{\mathrm{sin}\:^{\mathrm{2}} {t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:{A}\:=\:\int\frac{\mathrm{sin}\:{t}}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {t}}{dt} \\ $$$$\:\:\:\:\:\:\:{u}\:=\:\mathrm{cos}\:{t}\:\Rightarrow\:−{du}\:=\:\mathrm{sin}\:{tdt} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:=\:−\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} }\:\Rightarrow\:\:−\int\frac{{du}}{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:\:=\:\frac{−\mathrm{1}}{\mathrm{2}}\left\{\int\frac{{du}}{\mathrm{1}−{u}}+\int\:\frac{{du}}{\mathrm{1}+{u}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:=\:\frac{−\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{log}\:\left(\mathrm{1}−{u}\right)+\mathrm{log}\:\left(\mathrm{1}+{u}\right)\right\}\:+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\mid\:\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\:\mid+\:{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mid\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{1}+\mathrm{cos}\:{t}}\mid\:+{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mid\mathrm{tan}\:^{\mathrm{2}} \frac{{t}}{\mathrm{2}}\mid\:+\:{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:{A}\:=\:\mathrm{log}\:\mid\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\mid\:+\:{c} \\ $$

Facebook Tweet Pin

calculate-A-dt-sin-t-by-using-u-cos-t-

Leave a Reply Cancel reply