find-dx-1-tan-4-x-

Question Number 84569 by msup trace by abdo last updated on 14/Mar/20

$${find}\:\:\int\:\:\:\:\frac{{dx}}{\mathrm{1}+{tan}^{\mathrm{4}} {x}} \\ $$

Answered by M±th+et£s last updated on 14/Mar/20

∫((sec^2 (x))/((1+tan^2 (x))(1+tan^4 (x))))dx (1/2)∫((1+tan^4 (x)+1−tan^4 (x))/((1+tan^2 (x))(1+tan^4 (x)))) sec^2 (x) dx (1/2)∫((sec^2 (x))/((1+tan^2 (x))))dx +(1/2)∫((1−tan^2 (x))/((1+tan^4 (x))))sec^2 (x) dx (1/2)x+(1/2)∫((1−tan^2 (x))/(1+tan^4 (x)))sec^2 (x) (1/2)x+∫((((1/(tan^2 (x))) −1)sec^2 (x))/((tan(x)+(1/(tan(x))))−2)) dx (1/2)x+(1/( (√2)))∫(((1/( (√2)))(1−(1/(tan^2 (x))))sec^2 (x))/(1−(((tan(x)+(1/(tan(x))))/( (√2))))^2 ))dx (1/2)x+(1/( (√2)))tanh^(−1) (x)(((tan(x)+cot(x))/( (√2))))+c

$$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right)\left(\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)\right)}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)+\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right)\left(\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)\right)}\:{sec}^{\mathrm{2}} \left({x}\right)\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right)}{dx}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)\right)}{sec}^{\mathrm{2}} \left({x}\right)\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{tan}^{\mathrm{4}} \left({x}\right)}{sec}^{\mathrm{2}} \left({x}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}+\int\frac{\left(\frac{\mathrm{1}}{{tan}^{\mathrm{2}} \left({x}\right)}\:−\mathrm{1}\right){sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)+\frac{\mathrm{1}}{{tan}\left({x}\right)}\right)−\mathrm{2}}\:{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\mathrm{1}−\frac{\mathrm{1}}{{tan}^{\mathrm{2}} \left({x}\right)}\right){sec}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}−\left(\frac{{tan}\left({x}\right)+\frac{\mathrm{1}}{{tan}\left({x}\right)}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tanh}^{−\mathrm{1}} \left({x}\right)\left(\frac{{tan}\left({x}\right)+{cot}\left({x}\right)}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$$ \\ $$

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