1. Jnlucero

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

2. Jnlucero

find the value.

3. lgbasallote

have you tried using completing the square method?

4. wio

Looks like a candidate for trig substitution.

5. Jnlucero

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

6. lgbasallote

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?

7. wio

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

8. wio

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

9. Jnlucero

then what will be next?

10. wio

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

11. wio

Looks messy, but it really simplifies well.

12. Jnlucero

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

13. wio

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

14. wio

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

15. Jnlucero

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

16. wio

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

17. Jnlucero

ok......

18. Jnlucero

so how will i change it into $\sec \theta d \theta$