Question

integrate 9dx/1+9x^2

Answer 1

\[\Large \int\limits\limits\frac{9dx}{1+9x^2}\qquad=\qquad\int\limits\limits\frac{9dx}{1+(3x)^2}\]

We want to make the substitution, \(\large 3x=\tan\theta\)
Have you done much work with Trig subs yet?

Answer 2

Yes

Answer 3

\[\Large dx=\frac{1}{3}\sec^2\theta d \theta\]Applying the substitution gives us,\[\Large \int\limits\frac{9\left(\frac{1}{3}\sec^2\theta \;d \theta\right)}{1+\tan^2\theta}\]Understand how to simplify it down from there? :o

Answer 4

Do you use arctan?

Answer 5

After we integrate, yes arctan will be involved near the very end :)

Answer 6

But for now, it's just some simplification and then cancelling stuff out before we do an easy integration step.

Use your `Square Identity` in the denominator. \(\large 1+\tan^2\theta=?\)

Answer 7

1+tan^2=sec^2x?

Answer 8

\[\Large \int\limits\limits\frac{9\left(\frac{1}{3}\sec^2\theta \;d \theta\right)}{1+\tan^2\theta}\qquad=\qquad \int\limits\limits\frac{9\left(\frac{1}{3}\sec^2\theta \;d \theta\right)}{\sec^2\theta}\]Yesss good good.