Question

please help me on this

Answer 1

\[\int\limits_{0.5}^{1} \coth2x dx\] evaluate this

Answer 2

here is how to find the integral of cothx

http://math2.org/math/integrals/more/coth.htm

Answer 3

I get 0.5635 as the final answer :)

Answer 4

\[\int coth{2x} dx = \frac{1}{2}log{(sinh{2x})}\]

Answer 5

thanks :)

Answer 6

but i don't get the ln and modulus part.

Answer 7

okay so what part are you up to?

Answer 8

i get it when we use the exponent and u substitution,but what is the ln |u| + C ? integrated form of exponent?

Answer 9

yes so \[\int \frac{1}{u} du = ln(u)+c\]

My answer above was supposed to be a ln, sorry

Answer 10

alrighty now i get it :) thanks

Answer 11

no worries :)