Math, asked by namku, 1 year ago

find the solution of x :

Attachments:

namku: yes log2

namku: denominator inside root it is e^x

rational: and the right hand side is pi/6 ?

namku: yes

namku: m v.sorry for the unclear pic

rational: thats okay... evaluating the integral is easy by u substitution

namku: cant we use properties of definitive integral n substitute x+log2-x ie log2 in place of x ?

rational: thats a good idea, lets see...

rational: actually that wont work because the "x" in the bounds is a constant, which is "entirely different" from the variable of integration.

namku: ohhhhh i got it now awesome thanks

Answers

Answered by rational

substitute $u=\sqrt{e^x-1}\implies du~=~\frac{e^x}{2\sqrt{e^x-1}}dx$

(Notice that $u=\sqrt{e^x-1}\implies u^2=e^x-1\implies 1+u^2=e^x$ )

therefore the earlier differential becomes $du~=~\frac{1+u^2}{2\sqrt{e^x-1}}dx\implies \frac{2du}{1+u^2}=\frac{1}{\sqrt{e^x-1}}dx$ and the integral becomes

$=\int\limits_{1}^{\sqrt{e^x-1}}\frac{2du}{1+u^2}$

$=2\tan^{-1}(u)\Bigg|_{1}^{\sqrt{e^x-1}}$

$=2\tan^{-1}(\sqrt{e^x-1})-2\tan^{-1}(1)$

$=2\tan^{-1}(\sqrt{e^x-1})-\frac{\pi}{2}$

set that equal to $\frac{\pi}{6}$ and solve $x$ :
$2\tan^{-1}(\sqrt{e^x-1})-\frac{\pi}{2}~=~\frac{\pi}{6}$
$\tan^{-1}(\sqrt{e^x-1}~=~\frac{\pi}{3}$
$\sqrt{e^x-1}~=~\sqrt{3}$
$e^x-1~=~3$
$x~=~\log(4)$

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