Question Number 157929 by akolade last updated on 30/Oct/21
$$\int\frac{\mathrm{dx}}{\mathrm{sin}\:\mathrm{x}+\:\mathrm{sec}\:\mathrm{x}} \\ $$$$\mathrm{using}\:\mathrm{wiestress}\:\mathrm{substitution} \\ $$
Commented by cortano last updated on 30/Oct/21
![C = ∫ (1/( sin x+sec x)) dx [ tan (x/2) = u → { ((sin x =((2u)/(1+u^2 )))),((cos x=((1−u^2 )/(1+u^2 )))),((dx=(2/(1+u^2 )) du)) :}] C= ∫ (1/([((2u)/(1+u^2 ))+((1+u^2 )/(1−u^2 ))])) ((2/(1+u^2 )))du C=2∫ (((1−u^2 ))/(2u(1−u^2 )+(1+u^2 )^2 )) du C=2∫ ((1−u^2 )/(2u−2u^3 +u^4 +2u^2 +1)) du](https://www.tinkutara.com/question/Q158016.png)
$${C}\:=\:\int\:\frac{\mathrm{1}}{\:\mathrm{sin}\:{x}+\mathrm{sec}\:{x}}\:{dx}\: \\ $$$$\:\left[\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\:{u}\:\rightarrow\begin{cases}{\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\\{\mathrm{cos}\:{x}=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\\{{dx}=\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}}\end{cases}\right] \\ $$$${C}=\:\int\:\frac{\mathrm{1}}{\left[\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }+\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\mathrm{1}−{u}^{\mathrm{2}} }\right]}\:\left(\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$$${C}=\mathrm{2}\int\:\frac{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\mathrm{2}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)+\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du}\: \\ $$$${C}=\mathrm{2}\int\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{2}{u}−\mathrm{2}{u}^{\mathrm{3}} +{u}^{\mathrm{4}} +\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}}\:{du} \\ $$$$ \\ $$