(1)∫_e ^e^e ((ln (x).ln (ln (x)))/x) dx ? (2)lim_(x→π/4) ((cosec^2 x−2)/(cot x−1)) (3) Given { ((xy=((16y−9x)/(45)))),(((4/( (√x)))−(3/( (√y))) = 5)) :} ⇒find 9(√(xy))


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Question Number 110875 by bemath last updated on 31/Aug/20
$(1)∫_e ^e^e ((ln (x).ln (ln (x)))/x) dx ? (2)lim_(x→π/4) ((cosec^2 x−2)/(cot x−1)) (3) Given { ((xy=((16y−9x)/(45)))),(((4/( (√x)))−(3/( (√y))) = 5)) :} ⇒find 9(√(xy))$
$(1) \underset{e}{\int^{e^{e}}} \frac{\ln (x) . \ln (\ln (x))}{x} dx ?$ $(2) \lim_{x \to π / 4} \frac{cosec^{2} x - 2}{\cot x - 1}$ $(3) Given {\begin{cases} xy = \frac{16 y - 9 x}{45} \\ \frac{4}{\sqrt{x}} - \frac{3}{\sqrt{y}} = 5 \end{cases}$ $\Rightarrow find 9 \sqrt{xy}$
	Answered by Dwaipayan Shikari last updated on 31/Aug/20

	$\int_{e}^{e^{e}} \frac{l o g (x) l o g (l o g (x))}{x} d x (l o g x = t, \frac{1}{x} = \frac{d t}{d x})$ $\int_{1}^{e} t l o g (t) d t$ ${[l o g t . \frac{t^{2}}{2}]}_{1}^{e} - \int_{1}^{e} \frac{t}{2} = \frac{e^{2}}{2} - \frac{e^{2}}{4} + \frac{1}{4} = \frac{1}{4} (e^{2} + 1)$
	Commented by bemath last updated on 31/Aug/20

	$thank you sir .$
	Commented by bemath last updated on 31/Aug/20

	Answered by Dwaipayan Shikari last updated on 31/Aug/20

	$2) \lim_{x \to \frac{π}{4}} \frac{1 - 2 {s i n}^{2} x}{{s i n}^{2} x (c o s x - s i n x)} . s i n x = \lim_{x \to \frac{π}{4}} \frac{{c o s}^{2} x - {s i n}^{2} x}{s i n x (c o s x - s i n x)} = \sqrt{2} . (s i n x + c o s x)$ $= 2$
	Commented by bemath last updated on 31/Aug/20

	Answered by bobhans last updated on 31/Aug/20
	$(3) { ((45=((16y−9x)/(xy))=((16)/x)−(9/9)=((4/( (√x))))^2 −((3/( (√y))))^2 )),(((4/( (√x)))−(3/( (√y))) = 5)) :} { ((45=((4/( (√x)))−(3/( (√y))))((4/( (√x)))+(3/( (√y))))...(i))),((5=(4/( (√x)))−(3/( (√y)))...(ii))) :} ⇔ 45 = 5((4/( (√x)))+(3/( (√y)))) ; (4/( (√x)))+(3/( (√y))) = 9...(iii) (i)+(iii)⇒ (8/( (√x))) = 14⇒(√x) = (4/7) and (3/( (√y)))=9−7 (√y) = (3/2). therefore 9(√(xy)) = 9×(4/7)×(3/2) = ((54)/7)$
	$(3) {\begin{cases} 45 = \frac{16 y - 9 x}{xy} = \frac{16}{x} - \frac{9}{9} = {(\frac{4}{\sqrt{x}})}^{2} - {(\frac{3}{\sqrt{y}})}^{2} \\ \frac{4}{\sqrt{x}} - \frac{3}{\sqrt{y}} = 5 \end{cases}$ ${\begin{cases} 45 = (\frac{4}{\sqrt{x}} - \frac{3}{\sqrt{y}}) (\frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}}) . . . (i) \\ 5 = \frac{4}{\sqrt{x}} - \frac{3}{\sqrt{y}} . . . (ii) \end{cases}$ $\Leftrightarrow 45 = 5 (\frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}}); \frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 9 . . . (iii)$ $(i) + (iii) \Rightarrow \frac{8}{\sqrt{x}} = 14 \Rightarrow \sqrt{x} = \frac{4}{7} and \frac{3}{\sqrt{y}} = 9 - 7$ $\sqrt{y} = \frac{3}{2} . therefore 9 \sqrt{xy} = 9 \times \frac{4}{7} \times \frac{3}{2}$ $= \frac{54}{7}$
	Commented by bemath last updated on 31/Aug/20

	$thank you$
	Answered by mathmax by abdo last updated on 31/Aug/20
	$I =∫_e ^e^e ((ln(x)ln(lnx))/x) dx changement lnx =t give I =∫_1 ^e ((t ln(t))/e^t ) e^t dt =∫_1 ^e tln(t)dt =[(t^2 /2)ln(t)]_1 ^e −∫_1 ^e (t^2 /2)(dt/t) =(e^2 /2) −(1/2)∫_1 ^e t dt =(e^2 /2)−(1/2)[(t^2 /2)]_1 ^e =(e^2 /2)−(1/4){e^2 −1} =(e^2 /4) +(1/4) ⇒ I =((1+e^2 )/4)$
	$I = \int_{e}^{e^{e}} \frac{\ln (x) \ln (lnx)}{x} dx changement lnx = t give$ $I = \int_{1}^{e} \frac{t \ln (t)}{e^{t}} e^{t} dt = \int_{1}^{e} tln (t) dt = {[\frac{t^{2}}{2} \ln (t)]}_{1}^{e} - \int_{1}^{e} \frac{t^{2}}{2} \frac{dt}{t}$ $= \frac{e^{2}}{2} - \frac{1}{2} \int_{1}^{e} t dt = \frac{e^{2}}{2} - \frac{1}{2} {[\frac{t^{2}}{2}]}_{1}^{e} = \frac{e^{2}}{2} - \frac{1}{4} {e^{2} - 1} = \frac{e^{2}}{4} + \frac{1}{4} \Rightarrow$ $I = \frac{1 + e^{2}}{4}$
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