Question

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.. :/

Answer 1

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) ???

Answer 2

Well you don't want to use tan^2(theta) for u.

Answer 3

its tan(2theta) not... tan^2(theta) ...

Answer 4

And is your integral:
\[2 \int\limits \sinh(\tan(2\theta))\sec^2(2 \theta)dtheta\]?

Answer 5

Okay.

Answer 6

yes... I got to the part :
2cosh(tan(2(pi/8))) - (2cosh(tan(2(-pi/8))))

Answer 7

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:
\[\tan(-x) = -\tan(x); \implies \tan(- \frac{\pi}{4}) = -\tan(\frac{\pi}{4}) = -1\]

Answer 8

i cant understand your syntax... :/

Answer 9

Then the integration is easy. And if you want to simplify it even more replace cosh(u) (once you integrate) with:
\[\cosh(x) = \frac{e^x + e^{-x}}{2}\]

Answer 10

Your browser must not be updated if you can't see the latex properly.

Answer 11

im using internet explorer... :/

Answer 12

eff.. the answer is zero.. nvm.. thanks though

Answer 13

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.

Answer 14

http://www.wolframalpha.com/input/?i=plot+sinh%28x%29

Answer 15

thank you! :) knowledge!

Answer 16

while you're still here... what is the integral of cosh^2?