∫(1/(1+(√(cos(x))) )) dx


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Question Number 85414 by M±th+et£s last updated on 21/Mar/20

$\int \frac{1}{1 + \sqrt{c o s (x)}} d x$
Commented by M±th+et£s last updated on 21/Mar/20

$\int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt{c o s (x)}} d x t y p o . . . .$
	Answered by mind is power last updated on 23/Mar/20
	$∫_0 ^(π/2) (((1−(√(cos(x))))(1+cos(x)))/(1−cos^2 (x)))dx =∫_0 ^(π/2) ((2cos^2 ((x/2)))/(sin^2 (x))).(1−(√(cos(x))))dx =∫_0 ^(π/2) ((2cos^2 ((x/2))(1−(√(cos(x))))dx)/(4sin^2 ((x/2))cos^2 ((x/2)))) =∫_0 ^(π/2) (((1−(√(cos(x)))))/(2sin^2 ((x/2))))dx by part u′=(1/(2sin^2 ((x/2))))=(1/2)(cot^2 ((x/2))+1)⇒u=−cot((x/2)) =[−cot((x/2))(1−(√(cos(x))))]_0 ^(π/2) +∫_0 ^(π/2) ((cot((x/2)).sin(x))/(2(√(cos(x)))))dx =−1+∫_0 ^(π/2) ((cos((x/2))sin(x))/(sin((x/2)).2(√(cos(x))))) =−1+∫_0 ^(π/2) ((cos^2 ((x/2)))/(√(cos(x))))..A A=−1+∫_0 ^(π/2) ((1+cos(x))/(2(√(cos(x))))) =−1+∫_0 ^(π/2) ((((1/2)−sin((x/2))cos((x/2)))/(√(sin(x))))dx =−1+∫_0 ^(π/2) ((1+sin(x))/(2(√(sin(x)))))dx=−1+(1/2){∫_0 ^(π/2) (dx/(√(sin(x))))+∫_0 ^(π/2) (√(sin(x)))dx} ∫_0 ^(π/2) (dx/(√(sin(x)))) ∫_0 ^(π/2) (dx/(√(2sin((x/2))cos((x/2)))))=∫_0 ^(π/2) (dx/(√(1−(sin((x/2))−cos((x/2)))^2 ))) =∫_0 ^(π/2) (dx/(√(1−2sin^2 ((π/4)−(x/2)))))=∫_0 ^(π/4) ((2dr)/(√(1−2sin^2 (r))))=2F((π/4)∣2) ∫_0 ^(π/2) (√(sin(x)))dx=∫_0 ^(π/2) (√(1−(sin((x/2))−cos((x/2)))^2 ))dx =∫_0 ^(π/2) (√(1−2sin^2 ((π/4)−(x/2))))dx =2∫_0 ^(π/4) (√(1−2sin^2 (r)))dr=2E((π/4)∣2) =−1+2(E((π/4)∣2)+F((π/4)∣2))$
	$\int_{0}^{\frac{π}{2}} \frac{(1 - \sqrt{c o s (x)}) (1 + c o s (x))}{1 - {c o s}^{2} (x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{2 {c o s}^{2} (\frac{x}{2})}{{s i n}^{2} (x)} . (1 - \sqrt{c o s (x)}) d x$ $= \int_{0}^{\frac{π}{2}} \frac{2 {c o s}^{2} (\frac{x}{2}) (1 - \sqrt{c o s (x)}) d x}{4 {s i n}^{2} (\frac{x}{2}) {c o s}^{2} (\frac{x}{2})}$ $= \int_{0}^{\frac{π}{2}} \frac{(1 - \sqrt{c o s (x)})}{2 {s i n}^{2} (\frac{x}{2})} d x b y p a r t$ $u^{'} = \frac{1}{2 {s i n}^{2} (\frac{x}{2})} = \frac{1}{2} ({c o t}^{2} (\frac{x}{2}) + 1) \Rightarrow u = - c o t (\frac{x}{2})$ $= {[- c o t (\frac{x}{2}) (1 - \sqrt{c o s (x)})]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{c o t (\frac{x}{2}) . s i n (x)}{2 \sqrt{c o s (x)}} d x$ $= - 1 + \int_{0}^{\frac{π}{2}} \frac{c o s (\frac{x}{2}) s i n (x)}{s i n (\frac{x}{2}) . 2 \sqrt{c o s (x)}} = - 1 + \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} (\frac{x}{2})}{\sqrt{c o s (x)}} . . A$ $A = - 1 + \int_{0}^{\frac{π}{2}} \frac{1 + c o s (x)}{2 \sqrt{c o s (x)}}$ $= - 1 + \int_{0}^{\frac{π}{2}} \frac{(\frac{1}{2} - s i n (\frac{x}{2}) c o s (\frac{x}{2})}{\sqrt{s i n (x)}} d x$ $= - 1 + \int_{0}^{\frac{π}{2}} \frac{1 + s i n (x)}{2 \sqrt{s i n (x)}} d x = - 1 + \frac{1}{2} {\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n (x)}} + \int_{0}^{\frac{π}{2}} \sqrt{s i n (x)} d x}$ $\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n (x)}}$ $\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}} = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{1 - {(s i n (\frac{x}{2}) - c o s (\frac{x}{2}))}^{2}}}$ $= \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{1 - 2 {s i n}^{2} (\frac{π}{4} - \frac{x}{2})}} = \int_{0}^{\frac{π}{4}} \frac{2 d r}{\sqrt{1 - 2 {s i n}^{2} (r)}} = 2 F (\frac{π}{4} ∣ 2)$ $\int_{0}^{\frac{π}{2}} \sqrt{s i n (x)} d x = \int_{0}^{\frac{π}{2}} \sqrt{1 - {(s i n (\frac{x}{2}) - c o s (\frac{x}{2}))}^{2}} d x$ $= \int_{0}^{\frac{π}{2}} \sqrt{1 - 2 {s i n}^{2} (\frac{π}{4} - \frac{x}{2})} d x$ $= 2 \int_{0}^{\frac{π}{4}} \sqrt{1 - 2 {s i n}^{2} (r)} d r = 2 E (\frac{π}{4} ∣ 2)$ $= - 1 + 2 (E (\frac{π}{4} ∣ 2) + F (\frac{π}{4} ∣ 2))$
	Commented by M±th+et£s last updated on 23/Mar/20

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