1-e-e-e-ln-x-ln-ln-x-x-dx-2-lim-x-pi-4-cosec-2-x-2-cot-x-1-3-Given-xy-16y-9x-45-4-x-3-y-5-find-9-xy-

Question Number 110875 by bemath last updated on 31/Aug/20

(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) {\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}}) \dots (i) \\ 5 = \frac{4}{\sqrt{x}} - \frac{3}{\sqrt{y}} \dots (ii) \end{cases}

\Leftrightarrow 45 = 5 (\frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}}); \frac{4}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 9 \dots (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 = \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}