Submitted by jtedds on Fri, 03/17/2017 - 11:58
Forums:
I used the substitution u = coshx for the first part and similarly tried to use u = sinhx for the second part, but once I try to evaluate my integral I get the natural log of -1.
My integral is 1/(1 + 2u^2) which I evaluated to 1/sqr(2) * ln((u *sqr(2) + 1)/(u *sqr(2) - 1)) and my upper and lower bounds are sinha and 0 respectively
Note that $$\int_0^{\sinh a}
Note that $$\int_0^{\sinh a} \frac{1}{1+2u^2} \, \mathrm{d}u = \frac{1}{2}\int_0^{\sinh a} \frac{1}{\frac{1}{2} + u^2} \, \mathrm{d}u$$ is an $\arctan$ integral, not a $\log$ one. It would only be a $\log$ one if you have $1/2- u^2$ in the denominator.
Remember (this is in your formula booklet) that $\int \frac{\, \mathrm{d}x}{a^2 + x^2} = \frac{1}{a} \arctan \frac{x}{a} + \mathcal{c}$
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[That said, if you actually thought you got something that integrate to $\log$, remember to put modulus signs arorund the argument. So you would have $\displaystyle \bigg[\frac{1}{\sqrt{2}} \log \bigg| \frac{u \sqrt{2} + 1}{u\sqrt{2} -1}\bigg| \bigg]_0^{\sinh a}$ (note the moduli signs around the argument of the log) so the lower limit is actually $\frac{1}{\sqrt{2}} \ln |-1| = \frac{1}{\sqrt{2}} \ln (1) = 0$ (but of course, there's no logarithms in the second part, so this doesn't apply here)]
Thank you very much! A lesson
Thank you very much! A lesson in being careful with signs it seems, and thank you for the reminder when working with integrating to logs, I managed to struggle through the rest of the question.
Awesomesauce, glad it helped.
Awesomesauce, glad it helped.
The last part of Q1
Did anyone else find the last part of Q1 unclear? It says to sub $u=e^x$ in 'this result'. But after checking the solutions, it turns out it actually meant sub $u=e^x$ in 'the previous integral'. The 'result' would most obviously be $\frac{\pi}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$, but that doesn't have any $x$'s or $u$'s. So I started trying to sub into the $1+u^4$ integral, which got me nowhere.
Hopefully the step question setters read this comment and learn for next time.
I don't think so, assuming
I don't think so, assuming you're taking about 2004 III Q1 then the "result" is the entirety of this: $$\int_0^{\infty} \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d}x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}\ln \left(\frac{\sqrt{2} + 1}{\sqrt{2}-1}\right)$$
i.e: the result is the statement that the integral (LHS) equals the value (RHS), it's not just the value.
So substituting $u=e^x$ into the result very clearly refers to substituting it in the only viable place: the LHS, that is the integral.