Question Number 196487 by RoseAli last updated on 25/Aug/23
$$\int\left(\mathrm{sec}\:^{\mathrm{4}} {x}−\mathrm{cot}\:^{\mathrm{4}} {x}\right){dx} \\ $$
Answered by MM42 last updated on 26/Aug/23
![I_1 =∫sec^4 xdx=∫(1+tan^2 x)(1+tan^2 x)dx let tanx=u⇒ I_1 =∫(1+u^2 )du=u+(1/3)u^3 +c_1 =tanx+(1/3)tan^3 x+c_1 I_2 =∫cot^4 xdx=∫(cot^4 x+cot^2 x−cot^2 x−1+1)dx =∫ [cot^2 x(1+cot^2 x)−(1+cot^2 x)+1]dx =− (1/3)cot^3 x+cotx+x+c_2 ⇒I=tanx−cotx+(1/3)(tan^3 x+cot^3 x)−x+c ✓](https://www.tinkutara.com/question/Q196495.png)
$${I}_{\mathrm{1}} =\int{sec}^{\mathrm{4}} {xdx}=\int\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$${let}\:\:{tanx}={u}\Rightarrow \\ $$$${I}_{\mathrm{1}} =\int\left(\mathrm{1}+{u}^{\mathrm{2}} \right){du}={u}+\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} +{c}_{\mathrm{1}} ={tanx}+\frac{\mathrm{1}}{\mathrm{3}}{tan}^{\mathrm{3}} {x}+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int{cot}^{\mathrm{4}} {xdx}=\int\left({cot}^{\mathrm{4}} {x}+{cot}^{\mathrm{2}} {x}−{cot}^{\mathrm{2}} {x}−\mathrm{1}+\mathrm{1}\right){dx} \\ $$$$=\int\:\left[{cot}^{\mathrm{2}} {x}\left(\mathrm{1}+{cot}^{\mathrm{2}} {x}\right)−\left(\mathrm{1}+{cot}^{\mathrm{2}} {x}\right)+\mathrm{1}\right]{dx} \\ $$$$=−\:\frac{\mathrm{1}}{\mathrm{3}}{cot}^{\mathrm{3}} {x}+{cotx}+{x}+{c}_{\mathrm{2}} \\ $$$$\Rightarrow{I}={tanx}−{cotx}+\frac{\mathrm{1}}{\mathrm{3}}\left({tan}^{\mathrm{3}} {x}+{cot}^{\mathrm{3}} {x}\right)−{x}+{c}\:\checkmark \\ $$