dx-sin-x-sec-x-using-wiestress-substitution-

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 = \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

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