## anonymous 5 years ago integrate dx/9+4x^2

1. anonymous

Seems like you want a trig substitution here. Something like let $$2x = 3tan(\theta)$$

2. anonymous

polpak continue

3. anonymous

Well, what do you end up with using that substitution?

4. anonymous

$dx=3/2\sec ^{2}x thetad \theta$

5. anonymous

Yeah, though you have a stray x in there I'm assuming is a typo.

6. anonymous

So now convert your integral into the new form with respect to theta

7. anonymous

8. anonymous

Well, you have $\int \frac{1}{9+4x^2}dx$ Since 2x = 3tan$$\theta$$ What does that denominator become?

9. anonymous

$9\sec ^{2}\theta$

10. anonymous

$$9 + 9tan^2\theta$$

11. anonymous

Which yes, then becomes $$9sec^2\theta$$

12. anonymous

Ok, now you said dx = (3/2)$$sec^2\theta\ d\theta$$

13. anonymous

So we have $\frac{3}{2}\int\frac{sec^2\theta}{9sec^2\theta}d\theta$

14. anonymous

And I think you can take it from there.

15. anonymous

yea i got it thx.............

16. anonymous

polpak pls solve$\int\limits \tan ^{3}x$

17. anonymous

polpak i need ur help

18. anonymous

Hrm.. I don't recall the strategy for powers of tan I'm afraid. I'll see if I can re-derive it.

19. anonymous

ok

20. anonymous

integration by parts

21. anonymous

chaguanas i know it but solve it

22. anonymous

¬_¬ $\int \tan^3x = \int \tan x \ (\sec^2-1) = \int \tan x\ \sec^2x - \int \tan x$

23. anonymous

INewton pls continues

24. anonymous

For the first one, you could use a substitution of note it is of the form $f\ '(x) \cdot f \ (x)$ and try by recognition. $\int -\tan x = \int \frac{-\sin x}{\cos x}$ Which is of the form $\frac{f \ '(x)}{f\ (x)}$ (think ln )

25. anonymous

INewton pls solve $\int\limits \sqrt{9-4x ^{2}}/x$

26. anonymous

Same type deal as the other trig sub you had before except instead of tan you let 2x = $$3 sec\theta$$

27. anonymous

Lets see if you can do it yourself this time.