Question

Please help:

Answer 1

\[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]

Answer 2

find the value.

Answer 3

have you tried using completing the square method?

Answer 4

Looks like a candidate for trig substitution.

Answer 5

i do know the basic radical, but integrating radicals like that?.....oh thats confusing....

Answer 6

you should use completing the square first....that would give you an integral that looks like \[\int \sqrt{x^2 - a^2}dx\]
does that look familiar?

Answer 7

Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).

Answer 8

\[a=3,\ b=\sqrt{2}\]See what happens...

Answer 9

then what will be next?

Answer 10

Well if you did the substitution, you'd get
\[
x=\frac{3}{\sqrt{2}}\tan(\theta)\quad dx = \frac{3}{\sqrt{2}}\sec^2(\theta)d\theta
\]
\[ \Large
\int_1^2 \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2}\frac{3}{\sqrt{2}}\sec^2(\theta)d\theta
\]

Answer 11

Looks messy, but it really simplifies well.

Answer 12

looks messy. then what will be next.....

Answer 13

\[
\sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)}
\]

Answer 14

\[
= \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta)
\]

Answer 15

what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?

Answer 16

I was simplifying part of it. It's still there.

Answer 17

ok......

Answer 18

so how will i change it into \[\sec \theta d \theta\]