anonymous
  • anonymous
evaluate the following integral: 2sinh(tan2(theta))sec^2(2(theta)) d(theta) with limits of integration: (lower) -pi/8 , and (upper) pi/8 ..iv partially solved it with a u substitution but I dont know how to convert the limits of integration.. :/
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
so far i have 2cosh(u) .. with u = tan2(theta) ... soo it would be: 2cosh(tan(2(theta))) .. but idk what to do with the old limits of integration.. of (-pi/8 and pi/8) ???
anonymous
  • anonymous
Well you don't want to use tan^2(theta) for u.
anonymous
  • anonymous
its tan(2theta) not... tan^2(theta) ...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
And is your integral: \[2 \int\limits \sinh(\tan(2\theta))\sec^2(2 \theta)dtheta\]?
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
yes... I got to the part : 2cosh(tan(2(pi/8))) - (2cosh(tan(2(-pi/8))))
anonymous
  • anonymous
Well you need to evaluate: \[\tan(\pm 2 * \frac{\pi}{8})\] For the positive case you just have: \[\tan(\frac{\pi}{4}) = \frac{\sin\left(\frac{\pi}{4} \right)}{\cos \left( \frac{\pi}{4} \right)} = \frac{\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=1\] And since tangent is an odd function this means: \[\tan(-x) = -\tan(x); \implies \tan(- \frac{\pi}{4}) = -\tan(\frac{\pi}{4}) = -1\]
anonymous
  • anonymous
i cant understand your syntax... :/
anonymous
  • anonymous
Then the integration is easy. And if you want to simplify it even more replace cosh(u) (once you integrate) with: \[\cosh(x) = \frac{e^x + e^{-x}}{2}\]
anonymous
  • anonymous
Your browser must not be updated if you can't see the latex properly.
anonymous
  • anonymous
im using internet explorer... :/
anonymous
  • anonymous
eff.. the answer is zero.. nvm.. thanks though
anonymous
  • anonymous
Ah, this makes sense. Sinh(x) is an odd function which means that sinh(-x) = -sinh(x) so when you evaluate it over symmetric bounds it is zero.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=plot+sinh%28x%29
anonymous
  • anonymous
thank you! :) knowledge!
anonymous
  • anonymous
while you're still here... what is the integral of cosh^2?

Looking for something else?

Not the answer you are looking for? Search for more explanations.