Question

\[\int\limits \frac{dx}{x \sqrt{81x ^{2}-16}}\]

Answer 1

Trig Sub! I know how to do these now! :-D One moment, I'll show you steps... I need to get my lunch from the microwave oven :-3

Answer 2

First while I go to get this food, 81 = 9^2 and 16 = 4^2

Answer 3

Trig sub.

For this form with the 1/(x*sqrt(b^2*x^2-a^a))....

let \(x = \large\frac{4}{9}\sec\theta\)
\(dx = \large\frac{4}{9}\tan\theta\sec\theta\)

@Beatles watch what happens when you put this in place for "x" and "dx" :-)

Answer 4

\[\int\limits \frac{1}{x\cdot (9^2x^2-4^2)^{1/2}}\cdot dx\]
\[\int\limits \frac{1}{x\cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]

Answer 5

Take the unit circle identity for sine and cosine, divide both side by cos^2
\[\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}\]

Can you see what happens here, @Beatles? Can you write the identity here in terms of tangent-squared now?

Answer 6

Still there m8? :-D

Answer 7

lol ... the fastest way ... do opposite.

Answer 8

can we use the theorem of inverse trigo func. -_-

Answer 9

\[\int\limits \frac{1}{\sec\theta \cdot (9^2(\sec\theta)^2-4^2)^{1/2}}\cdot (\tan\theta\sec\theta \ \ d\theta)\]

@experimentX actually faster way is to use the integral tables ;-)

Answer 10

Err I mean wait that's 4/9 sec theta

Answer 11

Let me fix that...