1-1-sinx-dx-

Question Number 29885 by sinx last updated on 13/Feb/18

\int \frac{1}{1 + \sin x} d x = ?

Commented by MJS last updated on 14/Feb/18

= \tan x - \frac{1}{\cos x}

\dots sorry, {can}^{'} t show the way \dots

Commented by abdo imad last updated on 14/Feb/18

f (x) = \frac{t a n x c o s x - 1}{c o s x} = \frac{s i n x - 1}{c o s x} \Rightarrow

f^{^{'}} (x) = \frac{{c o s}^{2} x + (s i n x - 1) s i n x}{{c o s}^{2} x} = \frac{1 - s i n x}{1 - {s i n}^{2} x} = \frac{1}{1 + s i n x} s o

\int \frac{d x}{1 + s i n x} = t a n x - \frac{1}{c o s x} + k .

Answered by Joel578 last updated on 13/Feb/18

I = \int \frac{1}{1 + \sin x} . \frac{1 - \sin x}{1 - \sin x} d x

= \int \frac{1 - \sin x}{1 - \sin^{2} x} d x = \int \frac{1 - \sin x}{\cos^{2} x} d x

= \int (\frac{1}{\cos^{2} x} - \tan x \sec x) d x

= \int \sec^{2} x d x - \int \tan x \sec x d x

Commented by abdo imad last updated on 13/Feb/18

l e t p u t I = \int \frac{d x}{1 + s i n x} t h e c h . t a n (\frac{x}{2}) = t g i v e

I = \int \frac{1}{1 + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{1 + t^{2} + 2 t} = \int \frac{d t}{{(t + 1)}^{2}}

= - \frac{1}{t + 1} + k = \frac{- 1}{1 + t a n (\frac{x}{2})} + k

Commented by NECx last updated on 13/Feb/18

h m m m

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