Where did I go wrong? -integration

\displaystyle \int^a_0 \dfrac{sinh(x)}{2cosh^2(x)-1} \ dx

= \displaystyle \int^{cosh(a)}_{1}\dfrac{1}{2u^2-1} \ du

= -\dfrac{1}{\sqrt{2}} \displaystyle \int^{\sqrt{2}cosh(a)}_{\sqrt{2}} \dfrac{1}{1-k^2} \ dk

= -\dfrac{1}{\sqrt{2}} [artanh(k)]^{\sqrt{2}cosh(a)}_{\sqrt{2}}

= -\dfrac{1}{2\sqrt{2}} [log(\dfrac{1+k}{1-k})]^{\sqrt{2}cosh(a)}_{\sqrt{2}}

= \dfrac{1}{2\sqrt{2}}(log(\dfrac{1-\sqrt{2}cosh(a)}{1+\sqrt{2}cosh(a)}) + log(\dfrac{1+\sqrt{2}}{1-\sqrt{2}}))

Answer is meant to be:

= \dfrac{1}{2\sqrt{2}}(log(\dfrac{\sqrt{2}cosh(a)-1}{\sqrt{2}cosh(a)+1}) + log(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}))

Help is much appreciated! :kart:

(edited 13 years ago)

$Avatar for Get me off the £\?%!^@ computer$

Write the two log terms as a single log. Multiply top and bottom by -1 and then write as two separate logs again.

Original post by Get me off the £\?%!^@ computer

Write the two log terms as a single log. Multiply top and bottom by -1 and then write as two separate logs again.

Ah, I see! Thanks, so essentially, they're equivalent!

$Avatar for Get me off the £\?%!^@ computer$

Well, no. You're taking logs of negative numbers.

Original post by Get me off the £\?%!^@ computer

Well, no. You're taking logs of negative numbers.

Oh yeah, balls but surely something would cancel when using complex logarithms?

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