Question

HELP NEEDED.

Answer 1

Okay, so I have been stuck with this problem all day. My teacher put the same problem on the board, but just with a different coefficient... And I had solved it correctly then. Could you guys check to see if I did anything wrong please.

Answer 2

For some reason, WebAssign isn't accepting my answer.

Answer 3

i think that is good...

Answer 4

Yea... It's weird... Hmm

Answer 5

did you put the 9's in there?

Answer 6

What do you mean? There aren't any 9's in my answer. :3

Answer 7

\[9*\int\limits \sec^3(x)dx -9*\int\limits \sec(x)dx\]

should be your line 4 right?

Answer 8

oh i see, i think it looks good, not sure where the computer is not havin it on the way you type it in

Answer 9

Hm. Yea i'll ask my professor about it. Thanks anyway tho.

Answer 10

maybe it is just that constant, the last line should be integral of 9*sec(x)tan^2(x) to keep it the same as the initial problem

Answer 11

so you have 9/2 in front of the terms on the other side

Answer 12

\[\int\limits 9*\tan^2(x)*\sec(x)dx = \frac{ 9 }{ 2 }\sec(x)\tan(x) - \frac{ 9 }{ 2 }\ln (\sec(x)+\tan(x)) + C\]

try that

Answer 13

WOW. It worked..... Thank you so much!

Answer 14

no prob, yeah just forgot to keep it the same integral at the end ..

Answer 15

I see :)