The solution of the equation ∫_(log 2) ^x (1/(√(e^x −1))) dx= (π/6) is given by


Question and Answers Forum

All Questions Topic List
UNKNOWN Questions
Previous in All Question Next in All Question
Previous in UNKNOWN Next in UNKNOWN
Question Number 53697 by gunawan last updated on 25/Jan/19

$The solution of the equation$ $\underset{\log 2}{\int^{x}} \frac{1}{\sqrt{e^{x} - 1}} d x = \frac{π}{6} is given by$
Commented by Abdo msup. last updated on 25/Jan/19
$let A(x)=∫_(log2) ^x (dt/(√(e^t −1)))⇒A(x)=_(e^t =u) ∫_2 ^e^x (du/(u(√(u−1)))) =_((√(u−1))=α) ∫_1 ^(√(e^x −1)) ((2αdα)/((1+α^2 )α)) =2 ∫_1 ^(√(e^x −1)) (dα/(1+α^2 )) =2[arctan(α)]_1 ^(√(e^x −1)) =2{ arctan((√(e^x −1)))−(π/4)} =2arctan((√(e^x −1)))−(π/2) so A(x)=(π/6) ⇔2 arctan((√(e^x −1)))=(π/2) +(π/6) ⇒ 2 arctan((√(e^x −1)))=((4π)/6) =((2π)/3) ⇒ artan((√(e^x −1)))=(π/3) ⇒(√(e^x −1))=tan((π/3)) ⇒ (√(e^x −1))=(√3) ⇒e^x −1 =3 ⇒e^x =4 ⇒x =2ln(2) .$
$l e t A (x) = \int_{l o g 2}^{x} \frac{d t}{\sqrt{e^{t} - 1}} \Rightarrow A (x) =_{e^{t} = u} \int_{2}^{e^{x}} \frac{d u}{u \sqrt{u - 1}}$ $=_{\sqrt{u - 1} = α} \int_{1}^{\sqrt{e^{x} - 1}} \frac{2 α d α}{(1 + α^{2}) α} = 2 \int_{1}^{\sqrt{e^{x} - 1}} \frac{d α}{1 + α^{2}}$ $= 2 {[a r c t a n (α)]}_{1}^{\sqrt{e^{x} - 1}} = 2 {a r c t a n (\sqrt{e^{x} - 1}) - \frac{π}{4}}$ $= 2 a r c t a n (\sqrt{e^{x} - 1}) - \frac{π}{2} s o$ $A (x) = \frac{π}{6} \Leftrightarrow 2 a r c t a n (\sqrt{e^{x} - 1}) = \frac{π}{2} + \frac{π}{6} \Rightarrow$ $2 a r c t a n (\sqrt{e^{x} - 1}) = \frac{4 π}{6} = \frac{2 π}{3} \Rightarrow$ $a r t a n (\sqrt{e^{x} - 1}) = \frac{π}{3} \Rightarrow \sqrt{e^{x} - 1} = t a n (\frac{π}{3}) \Rightarrow$ $\sqrt{e^{x} - 1} = \sqrt{3} \Rightarrow e^{x} - 1 = 3 \Rightarrow e^{x} = 4 \Rightarrow x = 2 l n (2) .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum