How to solve this integral?

How to solve this integral?

Postby Guest » Wed Dec 11, 2019 2:02 pm

Can anyone help please?

Re: How to solve this integral?

Postby HallsofIvy » Fri Dec 13, 2019 8:21 pm

[tex]3^x= e^{ln(3^x)}= e^{x ln(3)}[/tex] and [tex]3^{-x}= e^{ln(3^{-x})}= e^{-x ln(3)}[/tex].
So [tex]\frac{1}{3^x+ 3^{-x}}=[/tex][tex]\frac{1}{e^{x ln(3)}+ e^{-x ln(3)}}=[/tex][tex]\frac{1}{2}\frac{2}{e^{x ln(3)}+ e^{-x ln(3)}}[/tex].

Now, cosh(x) is defined as [tex]\frac{e^x+ e^{-x}}{2}[/tex] so that is [tex]\frac{1}{2}\frac{1}{cosh(x ln(3))}[/tex]. The integral is [tex]\frac{1}{2}\int_{-1}^1 \frac{dx}{cosh(xln(3))}[/tex]

Letting [tex]u= x ln(3)[/tex], [tex]du= ln(3) dx[/tex], when x= 1, u= ln(3), and when x= -1, u= -ln(3) so that becomes [tex]\frac{1}{2 ln(3)}\int_{-ln(3)}^{ln(3)} \frac{du}{cosh(u)}[/tex].

That last integral is given at ... 721AAbWgg9

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