Question

integrate this one please

Answer 1

\[\int\limits_{}^{} 6x arcsec (3x) dx\]

Answer 2

@Spacelimbus

Answer 3

Just by intuition, use integration by parts and see what happens.

Answer 4

i know its by parts, but i got confused as i go through

Answer 5

\[ \large \int 6x \sec^{-1}(3x)dx = 3x^2\sec^{-1}(3x)-3 \int x^2 \cdot \frac{1}{x\sqrt{9x^2-1}} \]

Answer 6

So the second integrand becomes

\[ \large -3 \int \frac{x}{\sqrt{9x^2-1}} \] which looks like a trig substitution to me.

Answer 7

my answer is \[3x^2 \sin^{-1} (3x) -\frac{1}{3} \sqrt{9x^2-1}\]

Answer 8

i think i got it. :)

thanks though. :)

Answer 9

have you checked this answer?

Answer 10

it looks perfect, besides the sin^(-1) you mean sec^(-1)

Answer 11

Without trig substitution it's easier by the way, let u= 9x^2-1, but you have the answer already. So well done. (-:

Answer 12

oh, yes, it should be sec^-1

type it wrong again. haha

Answer 13

hehe, no prob, it's correct!