Question

hyperbolic function

medal

Answer 1

tan and a sec^2...
the integral is begging for u substitution : u = tan 6t

Answer 2

so sinh(tan(6t))1/6 cos^2(6 t)6

Answer 3

better formatting but yeh

Answer 4

then need to integrate the sec^2(6t)

Answer 5

so tan(6t)/6

Answer 6

\(\large u = \tan 6t \implies du = 6\sec^2 6t dt \implies \dfrac{du}{6} = \sec^2 6t dt\)

Answer 7

the indefinite integral becomes :

\[\large \int 9 \cosh u \dfrac{du}{6}\]

\[\large \dfrac{9}{6}\int \cosh u du\]

Answer 8

then sub in the sec^2(6t)

Answer 9

noticing that the given integrand is an even function may help you simplify the bounds, but its not so important for this problem..

Answer 10

after you int the cosh

Answer 11

so sinh

Answer 12

we did \(u = \tan 6t\) , right ?

Answer 13

yeah which turns into sec^2(6t)

Answer 14

so its (9/6)sinh(sec^2(6t))

Answer 15

not quite

Answer 16

\[\large \dfrac{9}{6}\int \cosh u du\]

\[\large \dfrac{9}{6} \sinh u\]


replace \(u\) by \(\tan 6t\), what do u get ?

Answer 17

ok i see it. then 9/6 sinh(pi/24)-9/(6sinh(-pi/24))

Answer 18

looks good ^^

Answer 19

bueno, thanks

Answer 20

we may simplify it a bit : sinh(-x) = -sinh(x)

Answer 21

9/6 sinh(pi/24)-9/6(sinh(-pi/24))

9/6 sinh(pi/24) + 9/6(sinh(pi/24))

3sinh(pi/24)