Question

Could someone help me integrate the attached? I have tried many ways but I still can't get the book answer of -pi/12

Answer 1

isnt that an inverse trig?

Answer 2

arcsec maybe looks familiar from the other day

Answer 3

maybe sec subitution

Answer 4

I tried the arcsec but I end up with an answer of -pi/8

Answer 5

or maybe factor out y^2 and pull it out of the radical in the denominator
and do a u sub : u = 1/y

Answer 6

nvm, it requires trig either way !

Answer 7

sites not posting stuff ...

y = asec(x)

y' = 1/(x sqrt(x^2-1))

so as long as the interval is within the appropriate interval ...

Answer 8

u = sqrt4x^2-1
works

Answer 9

it will simplify

Answer 10

I'm going to eat lunch. I'll try out the suggestions when I get back. Thanks all!

Answer 11

truust just do u=sqrt4y^2-1
after some algebra its simplfies very nicely

Answer 12

kx^2 - 1 = k(x^2 - 1/k)

1/(2x sqrt(x^2-1/4)) might be another way to look at it

Answer 13

ok nvm it goes back to arc tan anyway :)

Answer 14

u=sqrt(4y^2-1)
dy=u/8y du
integral u/u8y^2 = 1/32 integral 1/y^2 du = integral 1/(1+u^2) = 1/32 * arctan(u)
= 1/32 * arctan(4y^2-1)

Answer 15

there shud be a du on that first integral and 3rd integral xD

Answer 16

Hi all, thanks for your help! I was able to figure it out. I have attached it in case you were wondering how it was worked out.

Answer 17

@ear3232

Mathematica confirms your answer. Refer to the attachment.

Answer 18

Thank you, robtobey!