Question Number 158053 by Ar Brandon last updated on 30/Oct/21
$$\int\frac{{dx}}{\mathrm{sin}^{\mathrm{4}} {x}} \\ $$
Answered by ajfour last updated on 30/Oct/21
![∫(dx/([1−cos^2 x]^2 ))=∫(((1+t^2 )dt)/t^4 ) =−(1/(3t^3 ))−(1/t)+c =c−(((1+3tan^2 x)/(3tan^3 x))) .](https://www.tinkutara.com/question/Q158071.png)
$$\int\frac{{dx}}{\left[\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right]^{\mathrm{2}} }=\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}}{{t}^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{{t}}+{c} \\ $$$$={c}−\left(\frac{\mathrm{1}+\mathrm{3tan}\:^{\mathrm{2}} {x}}{\mathrm{3tan}\:^{\mathrm{3}} {x}}\right)\:. \\ $$
Answered by puissant last updated on 30/Oct/21
$$\Omega=\int\frac{\mathrm{1}}{{sin}^{\mathrm{4}} {x}}{dx}\:=\:\int\frac{{sec}^{\mathrm{4}} {x}}{{tan}^{\mathrm{4}} {x}}{dx}\: \\ $$$${u}={tanx}\:\rightarrow\:{du}={sec}^{\mathrm{2}} {xdx} \\ $$$$\Rightarrow\:\Omega\:=\:\int\frac{{sec}^{\mathrm{2}} {x}}{{u}^{\mathrm{4}} }{du}\:\Rightarrow\:\Omega\:=\:\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}^{\mathrm{4}} }{du} \\ $$$$=\:\int\left\{\frac{\mathrm{1}}{{u}^{\mathrm{4}} }+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right\}{du}\:=\:−\frac{\mathrm{1}}{\mathrm{3}{u}^{\mathrm{3}} }−\frac{\mathrm{1}}{{u}}+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\therefore\because\:\:\Omega\:=\:−\frac{\mathrm{1}}{\mathrm{3}}{cotan}^{\mathrm{3}} {x}−{cotanx}+{C} \\ $$