1. Integration? Hyperbolic?

$\displaystyle \int_{0}^{2} \frac{1}{\sqrt{256 +x^2}} dx$

Any suggestion on how to truck though this?

2. Originally Posted by Lolcats
$\displaystyle \int_{0}^{2} \frac{1}{\sqrt{256 +x^2}} dx$

Any suggestion on how to truck though this?
Try the substitution $\displaystyle x=16tan(\theta)$.

3. $\displaystyle \tan \theta=\dfrac{x}{16}$

4. Try the substitution .
Whats the reasoning behind this?

5. Oh nvm thanks Nacho!

6. k so I have $\displaystyle \displaystyle{\frac{1}{\sqrt{256+16\tan\!\left(u^{ 2}\right)}}\frac{du}{16\sec^{2}\!\left(u\right)}}$
where do I go from that?

Sorry if i dont respond right away, my laptops about to die and I have a chemistry seminar for the next hour just been trying to get some math done while I wait

7. You should get

$\displaystyle \int \frac{\sec^2 u * du}{\sqrt{256 + 256 \tan^2 u}}$

after the substitution.

8. Originally Posted by Lolcats
k so I have $\displaystyle \displaystyle{\frac{1}{\sqrt{256+16\tan\!\left(u^{ 2}\right)}}\frac{du}{16\sec^{2}\!\left(u\right)}}$
where do I go from that?
Be careful! $\displaystyle \left( {16\tan \theta } \right)^2 \ne 16\tan \left( {\theta ^2 } \right)$

And you donīt forget that: $\displaystyle 1+\tan^2 \theta=\sec^2 \theta$

9. You should get

after the substitution.
Isn't it

$\displaystyle \displaystyle{\frac{1}{\sqrt{256+256\tan^{2}\!\lef t(u\right)}}\!\left(16\sec^{2}\!\left(u\right)\rig ht)du}$

after the substitution?
Cuz the derivative of 16tan(u) = (16sec(u))^2
?

10. Now factor the denominator and use the identity $\displaystyle 1+ \tan^2 \theta = \sec^2 \theta$

11. 256=16^2 is the reasoning(correct me if i am wrong!).

12. k alright, after that I have...

$\displaystyle \displaystyle{\frac{\sec^{2}\!\left(u\right)du}{\s qrt{\sec^{2}\!\left(u\right)}}}$

13. i guess can you just sqrt the sec(U)^2? that would make this pretty easy then

14. $\displaystyle \displaystyle{\ln\!\left(\left|\tan\!\left(u\right )+\sec\!\left(u\right)\right|\right)}$
And now I'm utterly lost how to sub back u in... I would assume u=tan^(-1)(x)/(16)... but what do I do with something like that...?

This question looked so harmless to begin with...

15. Originally Posted by Lolcats
$\displaystyle \int_{0}^{2} \frac{1}{\sqrt{256 +x^2}} dx$

Any suggestion on how to truck though this?
Alternatively, try $\displaystyle x=16\sinh(x)$

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