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Calculus Level 2

Show that

∫ e x e 2 x + 1 d x ∫\frac{e^x}{e^{2x}+1} dx = arctan ⁡ ( e x ) \arctan (e^x) + C C ,

where C C is a constant of integration.

The volume of the solid formed by revolving the curve

f ( x ) f(x) = e x e 2 x + 1 \sqrt{\frac{e^x}{e^{2x}+1}}

bounded between x = l n 3 − ln ⁡ 3 x = ln\sqrt{3} - \ln{3} and x = 1 2 ln ⁡ 3 x = \frac{1}{2}\ln{3} around the x x axis can be expressed as π 2 a \frac{\pi ^2}{a} . Find the value of a a .

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The answer is 6.

Ethan Mandelez by Ethan Mandelez , 15, Malaysia

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2 solutions

Volume of Revolution \textrm{Volume of Revolution} , V = ∫ ζ Γ π ( f ( x ) ) 2 d x V=\int_ζ^{Γ} π (f(x))^2 dx = ∫ − 1 2 l o g ( 3 ) 1 2 l o g ( 3 ) π e x e 2 x + 1 d x =\int_{\frac{-1}{2}log(3)}^{\frac{1}{2}log(3)} π \frac{e^x}{e^{2x} +1}dx take e x = t e^x =t and the integral becomes ∫ 1 / √ 3 √ 3 π d t t 2 + 1 \int_{1/√3}^{√3} π\frac{dt}{t^2 +1} = π [ t a n − 1 t ] 1 / √ 3 √ 3 =π [tan^{-1} t ]^{√3}_{1/√3} = π ( π 3 − π 6 ) =π (\frac{π}{3} - \frac{π}{6}) = π 2 6 =\frac{π^2}{6}
EDIT🍎 \textrm{EDIT🍎} π 2 6 = 1 1 2 + 1 1 2 + 1 3 2 + 1 4 2 + ⋯ \frac{π^2}{6} = \frac{1}{1^2} + \frac{1}{1^2} + \frac{1}{3^2} +\frac{1}{4^2} + \cdots

2 Helpful 2 Interesting 2 Brilliant 0 Confused
Ethan Mandelez
Dec 2, 2020

1 Helpful 1 Interesting 1 Brilliant 0 Confused

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