help ! ! ! ∫ (dx/(csc(x)−1)) = ? [ my way ] ∫( (dx/((1/(sinx)) − 1)) ) =∫((sinx)/(1−sinx)) dx =−∫ ((sinx−1+1)/(sinx−1)) dx =−∫1+(1/(sinx−1)) dx =−(∫1dx+∫((sinx+1)/((sinx−1)(sinx+1))) dx) =−(x+C−∫((sinx+1)/(1−sin^2 x)) dx) =−(x+C−∫ ((sinx)/(cos^2 x)) dx−∫ (1/(cos^2 x)) dx) =−(x+C+∫(cosx)^(−2) dcosx−∫(1/(cos^2 x))dx) =−(x−(cosx)^(−1) +C−∫(1/(cos^2 x))dx) ...and I can′t solve the ∫(1/(cos^2 x))dx oh i just found that is tanx+C


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Question Number 33009 by kyle_TW last updated on 09/Apr/18

$h e l p!!!$ $\int \frac{d x}{c s c (x) - 1} = ?$ $[m y w a y]$ $\int (\frac{d x}{\frac{1}{s i n x} - 1})$ $= \int \frac{s i n x}{1 - s i n x} d x$ $= - \int \frac{s i n x - 1 + 1}{s i n x - 1} d x$ $= - \int 1 + \frac{1}{s i n x - 1} d x$ $= - (\int 1 d x + \int \frac{s i n x + 1}{(s i n x - 1) (s i n x + 1)} d x)$ $= - (x + C - \int \frac{s i n x + 1}{1 - {s i n}^{2} x} d x)$ $= - (x + C - \int \frac{s i n x}{{c o s}^{2} x} d x - \int \frac{1}{{c o s}^{2} x} d x)$ $= - (x + C + \int {(c o s x)}^{- 2} d c o s x - \int \frac{1}{{c o s}^{2} x} d x)$ $= - (x - {(c o s x)}^{- 1} + C - \int \frac{1}{{c o s}^{2} x} d x)$ $. . . a n d I {c a n}^{'} t s o l v e t h e \int \frac{1}{{c o s}^{2} x} d x$ $o h i j u s t f o u n d t h a t i s t a n x + C$
Commented by prof Abdo imad last updated on 09/Apr/18

$l e t p u t I = \int \frac{d x}{\frac{1}{s i n x} - 1}$ $I = \int \frac{s i n x}{1 - s i n x} d x$ $= - \int \frac{1 - s i n x - 1}{1 - s i n x} d x = - x + \int \frac{d x}{1 - s i n x}$ $t h e c h . t a n (\frac{x}{2}) = t g i v e$ $\int \frac{d x}{1 - s i n x} = \int \frac{1}{1 - \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int \frac{2 d t}{1 + t^{2} - 2 t} = 2 \int \frac{d t}{{(t - 1)}^{2}} = \frac{- 2}{t - 1} = \frac{2}{1 - t}$ $= \frac{2}{1 - t a n (\frac{x}{2})} \Rightarrow$ $I = - x + \frac{2}{1 - t a n (\frac{x}{2})} + λ .$
	Answered by Joel578 last updated on 09/Apr/18

	$I = \int \frac{d x}{(\frac{1}{\sin x} - \frac{\sin x}{\sin x})} = \int \frac{\sin x}{1 - \sin x} d x$ $= \int (\frac{\sin x}{1 - \sin x}) (\frac{1 + \sin x}{1 + \sin x}) d x$ $= \int \frac{\sin x + \sin^{2} x}{\cos^{2} x} d x$ $= \int \tan x \sec x d x + \int \tan^{2} x d x$ $= \int \tan x \sec x d x + \int \sec^{2} x - 1 d x$ $= \sec x + \tan x - x + C$
	Commented by kyle_TW last updated on 09/Apr/18

	$w o w t h a n k y o u v e r y m u c h$
	Commented by kyle_TW last updated on 09/Apr/18

	$I f o r g o t s e c x = \frac{1}{c o s x} . . .$
	Answered by math1967 last updated on 09/Apr/18

	$\int \frac{1}{c o s e x - 1} d x$ $\int \frac{c o s e x + 1}{{c o t}^{2} x} d x$ $\int \frac{c o s e c x}{{c o t}^{2} x} d x + \int {t a n}^{2} x d x$ $\int \frac{s i n x}{{c o s}^{2} x} d x + \int {s e c}^{2} x - \int d x$ $- \int \frac{d (c o s x)}{{c o s}^{2} x} + t a n x - x$ $\frac{1}{3} \times \frac{1}{{c o s}^{3} x} + t a n x - x + C$
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