∫cosecxdx


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Question Number 43158 by MASANJA J last updated on 07/Sep/18

$\int c o s e c x d x$
Commented by maxmathsup by imad last updated on 07/Sep/18

$l e t I = \int \frac{d x}{s i n (x)} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{1}{\frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{d t}{t} = l n ∣ t ∣ + c = l n ∣ t a n (\frac{x}{2}) ∣ + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18
	$∫(dx/(sinx)) ∫((sinx)/(sin^2 x))dx ∫((sinx)/(1−cos^2 x))dx t=cosx dt=−sinx dx ∫((−dt)/(1−t^2 )) ∫(dt/((t+1)(t−1))) (1/2)∫(((t+1)−(t−1))/((t+1)(t−1)))dt (1/2)[∫(dt/(t−1))−∫(dt/(t+1))] (1/2)[ln(t−1)−ln(t+1)] (1/2)ln∣((t−1)/(t+1))∣+c =(1/2)ln∣((cosx−1)/(cosx+1))∣+c =(1/2)ln∣((1−2sin^2 (x/2)−1)/(2cos^2 (x/2)))∣+c =(1/2)ln∣tan^2 (x/2)∣+c =ln(tan(x/2))+c recheck (d/dx){lntan(x/2)}=(1/(tan(x/2)))×sec^2 (x/2)×(1/2) =(1/(2×((sin(x/2))/(cos(x/2)))×cos^2 (x/2))) =(1/(2sin(x/2)×cos(x/2)))=(1/(sinx))=cosecx$
	$\int \frac{d x}{s i n x}$ $\int \frac{s i n x}{{s i n}^{2} x} d x$ $\int \frac{s i n x}{1 - {c o s}^{2} x} d x$ $t = c o s x d t = - s i n x d x$ $\int \frac{- d t}{1 - t^{2}}$ $\int \frac{d t}{(t + 1) (t - 1)}$ $\frac{1}{2} \int \frac{(t + 1) - (t - 1)}{(t + 1) (t - 1)} d t$ $\frac{1}{2} [\int \frac{d t}{t - 1} - \int \frac{d t}{t + 1}]$ $\frac{1}{2} [l n (t - 1) - l n (t + 1)]$ $\frac{1}{2} l n ∣ \frac{t - 1}{t + 1} ∣ + c$ $= \frac{1}{2} l n ∣ \frac{c o s x - 1}{c o s x + 1} ∣ + c$ $= \frac{1}{2} l n ∣ \frac{1 - 2 {s i n}^{2} \frac{x}{2} - 1}{2 {c o s}^{2} \frac{x}{2}} ∣ + c$ $= \frac{1}{2} l n ∣ {t a n}^{2} \frac{x}{2} ∣ + c$ $= l n (t a n \frac{x}{2}) + c$ $r e c h e c k$ $\frac{d}{d x} {l n t a n \frac{x}{2}} = \frac{1}{t a n \frac{x}{2}} \times {s e c}^{2} \frac{x}{2} \times \frac{1}{2}$ $= \frac{1}{2 \times \frac{s i n \frac{x}{2}}{c o s \frac{x}{2}} \times {c o s}^{2} \frac{x}{2}}$ $= \frac{1}{2 s i n \frac{x}{2} \times c o s \frac{x}{2}} = \frac{1}{s i n x} = c o s e c x$
	Answered by $@ty@m last updated on 08/Sep/18

	$\int cosec x \times \frac{cosec x + \cot x}{cosec x + \cot x} d x$ $= - \int \frac{- cosec^{2} x - cosec x . \cot x}{cosec x + \cot x} d x$ $= - \ln (cosec x + \cot x) + C$
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