∫ (dx/(sin x (√(cos x)))) =?


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Question Number 136365 by liberty last updated on 21/Mar/21

$\int \frac{d x}{\sin x \sqrt{\cos x}} = ?$
	Answered by mathmax by abdo last updated on 21/Mar/21
	$I =∫ (dx/(sinx(√(cosx)))) we do the changement cosx=t^2 ⇒x=arcos(t^2 ) ⇒sinx=(√(1−t^4 )) ⇒I =∫ ((−2t)/( (√(1−t^4 )).(√(1−t^4 ))t))dt =−2∫ (dt/(1−t^4 )) =2∫ (dt/(t^4 −1)) let decompose F(t)=(1/(t^4 −1)) ⇒F(t)=(1/((t^2 −1)(t^2 +1))) =(1/((t−1)(t+1)(t^2 +1)))=(a/(t−1))+(b/(t+1)) +((nt+m)/(t^2 +1)) a=(1/4) ,b=−(1/4) , lim_(t→+∞) tF(t)=0 =a+b+n ⇒n=0 F(o)=−1=−a+b +m ⇒m=−1+a−b =−1+(1/2)=−(1/2) ⇒ F(t)=(1/(4(t−1)))−(1/(4(t+1)))−(1/(2(t^2 +1))) ⇒ I =−2∫ F(t)dt =−2{(1/4)ln∣((t−1)/(t+1))∣−(1/2) arctan(t)} +C =−(1/2)ln∣((t−1)/(t+1))∣+arctan(t) +C =arctan((√(cosx)))−(1/2)ln∣(((√(cosx))−1)/( (√(cosx))+1))∣ +C$
	$I = \int \frac{dx}{sinx \sqrt{cosx}} we do the changement cosx = t^{2} \Rightarrow x = arcos (t^{2})$ $\Rightarrow sinx = \sqrt{1 - t^{4}} \Rightarrow I = \int \frac{- 2 t}{\sqrt{1 - t^{4}} . \sqrt{1 - t^{4}} t} dt = - 2 \int \frac{dt}{1 - t^{4}}$ $= 2 \int \frac{dt}{t^{4} - 1} let decompose F (t) = \frac{1}{t^{4} - 1} \Rightarrow F (t) = \frac{1}{(t^{2} - 1) (t^{2} + 1)}$ $= \frac{1}{(t - 1) (t + 1) (t^{2} + 1)} = \frac{a}{t - 1} + \frac{b}{t + 1} + \frac{nt + m}{t^{2} + 1}$ $a = \frac{1}{4}, b = - \frac{1}{4}, \lim_{t \to + \infty} tF (t) = 0 = a + b + n \Rightarrow n = 0$ $F (o) = - 1 = - a + b + m \Rightarrow m = - 1 + a - b = - 1 + \frac{1}{2} = - \frac{1}{2} \Rightarrow$ $F (t) = \frac{1}{4 (t - 1)} - \frac{1}{4 (t + 1)} - \frac{1}{2 (t^{2} + 1)} \Rightarrow$ $I = - 2 \int F (t) dt = - 2 {\frac{1}{4} \ln ∣ \frac{t - 1}{t + 1} ∣ - \frac{1}{2} \arctan (t)} + C$ $= - \frac{1}{2} \ln ∣ \frac{t - 1}{t + 1} ∣ + \arctan (t) + C$ $= \arctan (\sqrt{cosx}) - \frac{1}{2} \ln ∣ \frac{\sqrt{cosx} - 1}{\sqrt{cosx} + 1} ∣ + C$
	Answered by EDWIN88 last updated on 21/Mar/21
	$I = ∫ (dx/(sin x (√(cos x)))) = ∫ ((sin x)/(sin^2 x (√(cos x)))) dx I=∫((−d(cos x))/((1−cos^2 x)(√(cos x)))) = −∫ (du/((1−u^2 )(√u))) I=∫ (du/((u^2 −1)(√u))) = ∫ (du/((u−1)(u+1)(√u))) let (√u) = s⇒u =s^2 , du = 2s ds I=∫ ((2s)/((s^2 −1)(s^2 +1)s)) ds = ∫ ((2ds)/((s+1)(s−1)(s^2 +1))) Partial fraction (2/((s+1)(s−1)(s^2 +1)))=(a/(s+1))+(b/(s−1))+((cs+d)/(s^2 +1))$
	$I = \int \frac{dx}{\sin x \sqrt{\cos x}} = \int \frac{\sin x}{\sin^{2} x \sqrt{\cos x}} dx$ $I = \int \frac{- d (\cos x)}{(1 - \cos^{2} x) \sqrt{\cos x}} = - \int \frac{du}{(1 - u^{2}) \sqrt{u}}$ $I = \int \frac{du}{(u^{2} - 1) \sqrt{u}} = \int \frac{du}{(u - 1) (u + 1) \sqrt{u}}$ $let \sqrt{u} = s \Rightarrow u = s^{2}, du = 2 s ds$ $I = \int \frac{2 s}{(s^{2} - 1) (s^{2} + 1) s} ds = \int \frac{2 ds}{(s + 1) (s - 1) (s^{2} + 1)}$ $Partial fraction \frac{2}{(s + 1) (s - 1) (s^{2} + 1)} = \frac{a}{s + 1} + \frac{b}{s - 1} + \frac{cs + d}{s^{2} + 1}$
	Answered by mindispower last updated on 21/Mar/21

	$= \int \frac{s i n (x) d x}{1 - {c o s}^{2} (x)} . \frac{1}{\sqrt{c o s (x)}}$ $u = \sqrt{c o s (x)}$ $\Leftrightarrow - 2 \int \frac{d u}{(1 - u^{4})}$ $= \int (\frac{1}{u^{2} - 1} + \frac{1}{u^{2} + 1}) d u$ $= \frac{1}{2} l n ∣ \frac{u - 1}{u + 1} ∣ + \tan^{- 1} (u) + c$ $= l n (\sqrt{\frac{1 - \sqrt{c o s (x)}}{1 + \sqrt{c o s (x)}})}) + {t a n}^{-} (\sqrt{c o s (x)}) + c$
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