How do you find the integral of $\int \frac{1}{1 + sin (x)}$ $int 1/(1 + sin(x))$ ?

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How do you find the integral of $\int \frac{1}{1 + sin (x)}$ $int 1/(1 + sin(x))$ ?

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Answer 1 · 2015-10-08T05:03:15

Apply the following substitution

$u = tan (\frac{x}{2})$ $u=tan(x/2)$ hence $d x = \frac{2}{1 + u^{2}} d u$ $dx=2/(1+u^2)du$ , $sin x = \frac{2 u}{1 + u^{2}}$ $sinx=(2u)/(1+u^2)$

hence you get the integral

$\int \frac{1}{1 + sin x} d x = \int \frac{2}{{(u + 1)}^{2}} d u = - \frac{2}{1 + u} + c = - \frac{2}{1 + tan (\frac{x}{2})} + c$ $int 1/(1+sinx)dx=int 2/(u+1)^2du=-2/(1+u)+c=-2/(1+tan(x/2))+c$

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How do you find the integral of $\int \frac{1}{1 + sin (x)}$ $int 1/(1 + sin(x))$ ?

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