Question Number 86397 by M±th+et£s last updated on 28/Mar/20
$$\int\frac{{dx}}{{sin}^{\mathrm{2}} \left({x}\right)+{tan}^{\mathrm{2}} \left({x}\right)}\:{dx} \\ $$
Commented by Prithwish Sen 1 last updated on 28/Mar/20
$$−\mathrm{cotx}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{tan}^{\mathrm{2}} \mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mid+\mathrm{C} \\ $$$$\mathrm{C}=\:\mathrm{arbitrary}\:\mathrm{constant} \\ $$
Commented by jagoll last updated on 28/Mar/20
$$\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} \:\mathrm{x}\:\mathrm{dx}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}}{\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$\int\:\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)}−\int\frac{\mathrm{dx}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{csc}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:−\int\:\frac{\mathrm{dx}}{\mathrm{3}+\mathrm{cos}\:\mathrm{2x}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:\mathrm{x}\:−\int\:\frac{\mathrm{dx}}{\mathrm{3}+\mathrm{cos}\:\mathrm{2x}} \\ $$$$\mathrm{next} \\ $$
Answered by Prithwish Sen 1 last updated on 28/Mar/20
![2∫(dx/((1−cos2x)+2(((1−cos2x))/((1+cos2x))))) =2∫(((1+cos2x)dx)/((1−cos2x)(3−cos2x))) =2[∫(1/(1−cos2x)) −(2/(3−cos2x))]dx =∫coec^2 x dx −2∫((sec^2 x)/((1+2tan^2 x))) dx =−cotx−(1/2)ln∣tan^2 x +(1/2)∣ +C please check.](https://www.tinkutara.com/question/Q86402.png)
$$\mathrm{2}\int\frac{\mathrm{dx}}{\left(\mathrm{1}−\mathrm{cos2x}\right)+\mathrm{2}\frac{\left(\mathrm{1}−\mathrm{cos2x}\right)}{\left(\mathrm{1}+\mathrm{cos2x}\right)}} \\ $$$$=\mathrm{2}\int\frac{\left(\mathrm{1}+\mathrm{cos2x}\right)\mathrm{dx}}{\left(\mathrm{1}−\mathrm{cos2x}\right)\left(\mathrm{3}−\mathrm{cos2x}\right)} \\ $$$$=\mathrm{2}\left[\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos2x}}\:−\frac{\mathrm{2}}{\mathrm{3}−\mathrm{cos2x}}\right]\mathrm{dx} \\ $$$$=\int\mathrm{coec}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:−\mathrm{2}\int\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}}{\left(\mathrm{1}+\mathrm{2tan}^{\mathrm{2}} \mathrm{x}\right)}\:\mathrm{dx} \\ $$$$=−\boldsymbol{\mathrm{cotx}}−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mid\:+\boldsymbol{\mathrm{C}} \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Answered by MJS last updated on 28/Mar/20
![∫(dx/(sin^2 x +tan^2 x))= [t=tan x → dx=cos^2 x dt] =∫(dt/(t^2 (t^2 +2)))=(1/2)∫(dt/t^2 )−(1/2)∫(dt/(t^2 +2))= =−(1/(2t))−((√2)/4)arctan (((√2)t)/2) = =−(1/(2tan x))−((√2)/4)arctan (((√2)tan x)/2) +C](https://www.tinkutara.com/question/Q86419.png)
$$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{2}} \:{x}\:+\mathrm{tan}^{\mathrm{2}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{t}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}{t}}{\mathrm{2}}\:= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2tan}\:{x}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:{x}}{\mathrm{2}}\:+{C} \\ $$
Commented by M±th+et£s last updated on 28/Mar/20
$${thank}\:{you}\:{sir} \\ $$