Question Number 216772 by Nadirhashim last updated on 19/Feb/25
$$\:\:\boldsymbol{{find}}\:\int\frac{\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{sec}}^{\mathrm{4}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\:\: \\ $$
Answered by MrGaster last updated on 19/Feb/25
![∫((tan^2 (x) )/(1−sec^4 (x))) .dx =∫((tan^2 (x))/(1−((1/(cos^2 (x))))^2 ))dx=∫((tan^2 (x))/((cos^4 (x)−1)/(cos^4 (x))))dx=∫((tan^2 (x)cos^4 (x))/(cos^4 (x)−1))dx =∫((sin^2 (x))/(cos^2 (x)(cos^4 (x)−1)))dx=∫((sin^2 (x))/(cos^2 (x)(cos^2 (x)−1)(cos^2 (x)+1))))dx =∫((sin^2 (x))/(cos^2 (x)(−sin^2 (x)(cos^2 (x)+1))))dx=−∫((sin^2 (x))/(sin^2 (x)(cos^2 (x)+1)))dx =∫(1/(cos^2 (x)(cos^2 (x)+1)))dx−∫((sec^2 (x))/(1+cos^2 (x)))dx =−∫((sec^2 (x))/(2−sin^2 (x)))dx Let u=tan(x)⇒du=sec^2 (x)dx. =∫(1/(2−u^2 ))du=−(1/( (√2)))∫(1/( (√2)−u))+(1/( (√2)+u))du =−(1/( (√2)))[ln∣(√2)−u∣−ln∣(√2)+u∣]+C =−(1/2)ln∣(((√2)−tan(x))/( (√2)+tan(x)))∣+C =(1/2)ln∣(((√2)+tan(x))/( (√2)−tan(x)))∣+C =((tan^(−1) (((tan(x))/( (√2)))))/( (√2)))−x+C](https://www.tinkutara.com/question/Q216775.png)
$$\:\int\frac{\boldsymbol{{tan}}^{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:}{\mathrm{1}−\boldsymbol{{sec}}^{\mathrm{4}} \left(\boldsymbol{{x}}\right)}\:.\boldsymbol{{dx}}\:=\int\frac{\mathrm{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left({x}\right)}\right)^{\mathrm{2}} }{dx}=\int\frac{\mathrm{tan}^{\mathrm{2}} \left({x}\right)}{\frac{\mathrm{cos}^{\mathrm{4}} \left({x}\right)−\mathrm{1}}{\mathrm{cos}^{\mathrm{4}} \left({x}\right)}}{dx}=\int\frac{\mathrm{tan}^{\mathrm{2}} \left({x}\right)\mathrm{cos}^{\mathrm{4}} \left({x}\right)}{\mathrm{cos}^{\mathrm{4}} \left({x}\right)−\mathrm{1}}{dx} \\ $$$$=\int\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{cos}^{\mathrm{2}} \left({x}\right)\left(\mathrm{cos}^{\mathrm{4}} \left({x}\right)−\mathrm{1}\right)}{dx}=\int\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{\left.\mathrm{cos}^{\mathrm{2}} \left({x}\right)\left(\mathrm{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\right)\left(\mathrm{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)\right)}{dx} \\ $$$$=\int\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{cos}^{\mathrm{2}} \left({x}\right)\left(−\mathrm{sin}^{\mathrm{2}} \left({x}\right)\left(\mathrm{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)\right)}{dx}=−\int\frac{\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{sin}^{\mathrm{2}} \left({x}\right)\left(\mathrm{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left({x}\right)\left(\mathrm{cos}^{\mathrm{2}} \left({x}\right)+\mathrm{1}\right)}{dx}−\int\frac{\mathrm{sec}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$=−\int\frac{\mathrm{sec}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\mathrm{Let}\:{u}=\mathrm{tan}\left({x}\right)\Rightarrow{du}=\mathrm{sec}^{\mathrm{2}} \left({x}\right){dx}. \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{2}−{u}^{\mathrm{2}} }{du}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−{u}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+{u}}{du} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{ln}\mid\sqrt{\mathrm{2}}−{u}\mid−\mathrm{ln}\mid\sqrt{\mathrm{2}}+{u}\mid\right]+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{tan}\left({x}\right)}{\:\sqrt{\mathrm{2}}+\mathrm{tan}\left({x}\right)}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{2}}+\mathrm{tan}\left({x}\right)}{\:\sqrt{\mathrm{2}}−\mathrm{tan}\left({x}\right)}\mid+{C} \\ $$$$=\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\left({x}\right)}{\:\sqrt{\mathrm{2}}}\right)}{\:\sqrt{\mathrm{2}}}−{x}+{C} \\ $$
Answered by Fridayja last updated on 21/Jul/25
$${wrong} \\ $$