∫(dx/(sin x+ sec x)) using wiestress substitution


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 157929 by akolade last updated on 30/Oct/21

$\int \frac{dx}{\sin x + \sec x}$ $using wiestress substitution$
Commented by cortano last updated on 30/Oct/21
$C = ∫ (1/( sin x+sec x)) dx [ tan (x/2) = u → { ((sin x =((2u)/(1+u^2 )))),((cos x=((1−u^2 )/(1+u^2 )))),((dx=(2/(1+u^2 )) du)) :}] C= ∫ (1/([((2u)/(1+u^2 ))+((1+u^2 )/(1−u^2 ))])) ((2/(1+u^2 )))du C=2∫ (((1−u^2 ))/(2u(1−u^2 )+(1+u^2 )^2 )) du C=2∫ ((1−u^2 )/(2u−2u^3 +u^4 +2u^2 +1)) du$
$C = \int \frac{1}{\sin x + \sec x} d x$ $[\tan \frac{x}{2} = u \to {\begin{cases} \sin x = \frac{2 u}{1 + u^{2}} \\ \cos x = \frac{1 - u^{2}}{1 + u^{2}} \\ d x = \frac{2}{1 + u^{2}} d u \end{cases}]$ $C = \int \frac{1}{[\frac{2 u}{1 + u^{2}} + \frac{1 + u^{2}}{1 - u^{2}}]} (\frac{2}{1 + u^{2}}) d u$ $C = 2 \int \frac{(1 - u^{2})}{2 u (1 - u^{2}) + {(1 + u^{2})}^{2}} d u$ $C = 2 \int \frac{1 - u^{2}}{2 u - 2 u^{3} + u^{4} + 2 u^{2} + 1} d u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum