Question

integrate 5x^3/(x^2+9)^(1/2) using trig sub

Answer 1

x=3tan(theta)

Answer 2

dx=3sec^2(theta)

Answer 3

seems good so far, if that's not doing the trick, you might want to try a hyperbolic trigonometric substitution.

Answer 4

Now just tack on that sqrt(x^2+9) = 3sec(theta)

Answer 5

\[5\int\limits_{}^{}\frac{ (3\tan(\theta)(3\sec ^{3}(\theta) }{3\sec(\theta) }\]

Answer 6

\[45\int\limits_{}^{}\tan(\theta)\sec ^{2}(\theta)\]

Answer 7

oh

Answer 8

I just realized one of my mistakes

Answer 9

Looks like youre answering your own question xD

Answer 10

haha, it should only be 15 since two of the 3's cancel

Answer 11

Didnt check, you were doing too well on your own, lol.

Answer 12

So I have \[15\int\limits_{}^{}\sec(\theta)\tan(\theta)\sec(\theta)\]

Answer 13

\[u=\sec(\theta) du=\sec(\theta)\tan(\theta)\]

Answer 14

\[15\int\limits_{}^{}u du\]

Answer 15

\[(\frac{ 15 }{ 1 })(\frac{ 1 }{ 2 })u ^{2}\]

Answer 16

\[(\frac{ 15 }{ 2 })\sec ^{2}(\theta)\]

Answer 17

\[\sec(\theta) = \frac{ \sqrt{x ^{2}+9} }{ 3 }\]

Answer 18

\[\frac{ 15\sqrt{x ^{2}+9} }{ 18 }\]

Answer 19

Did I do something wrong? It says I'm getting the wrong answer

Answer 20

You didnt square the square root portion.

Answer 21

ohhhh, lol

Answer 22

I just eliminated the square root on top and it's still counting it wrong

Answer 23

Alright, ill work it out then.

Answer 24

any luck?

Answer 25

Yeah, I got it now, im just being dumb, lol x_x

Answer 26

haha, nice

Answer 27

Shouldve had this after your substitutions;