Question

Help please!! Trigonometric substitution
integrate 1/(x^5*sqrt((9x^2-1))
on interval [sqrt(2)/3,2/3]

Answer 1

You wanna use \(x=\sec(\theta)/3\)

Answer 2

thanks for responding.. why is it 1/3 and not just 3?

Answer 3

Remember:\[
b^2x+a^2 \to x = \frac{a}{b}\tan\theta \\
b^2x-a^2 \to x = \frac{a}{b}\sec\theta \\
a^2-b^2x \to x = \frac{a}{b}\sin\theta
\]

Answer 4

In this case \(b^2=9\) and \(a^2=1\), this means \(a/b = 1/3\)

Answer 5

You want to have \((1/3)^2\) to cancel out the \(9\) coefficient next to \(x\).

Answer 6

Ohhh, that I didn't know. Thank you.

Answer 7

and then my boundaries would change to pi/18 and ??

Answer 8

Well remember that \[
x = \frac{\sec(\theta)}{3} = \frac{1}{3\cos(\theta)}\implies \frac{1}{\cos(\theta)} = 3x\implies \cos\theta = \frac{1}{3x}
\]

Answer 9

So you have \(\cos(\theta_1) =1/\sqrt{2} \), \(\cos(\theta_2) =1/2 \)

Answer 10

These are fairly common angles.

Answer 11

ok, i'll give it another shot. thanks for your help

Answer 12

Got it! Really appreciate your help!