Question

integrate dx/9+4x^2

Answer 1

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

Answer 2

polpak continue

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

please help idk

Answer 8

Well, you have
\[\int \frac{1}{9+4x^2}dx\]

Since 2x = 3tan\(\theta\)
What does that denominator become?

Answer 9

\[9\sec ^{2}\theta\]

Answer 10

\(9 + 9tan^2\theta\)

Answer 11

Which yes, then becomes \(9sec^2\theta\)

Answer 12

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

Answer 13

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

Answer 14

And I think you can take it from there.

Answer 15

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

Answer 16

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

Answer 17

polpak i need ur help

Answer 18

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

Answer 19

ok

Answer 20

integration by parts

Answer 21

chaguanas i know it but solve it

Answer 22

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

Answer 23

INewton pls continues

Answer 24

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 )

Answer 25

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

Answer 26

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

Answer 27

Lets see if you can do it yourself this time.