inverse trig integral: [int] 1/(9-81x^2)^(1/2) dx

[dmsjr12]

I found this problem in my homework and was looking for any help I could get with it. I know I have to use substitution and the sin function, but I'm not completely sure how to do it. Any help would be appreciated.

[int] 1/(9-81x^2)^(1/2) dx

 

[royhaas]

Use the substitution \(\displaystyle 9x=3sin(u)\).

 

[galactus]

\(\displaystyle \L\\\int\frac{1}{\sqrt{9-81x^{2}}}dx\)

\(\displaystyle \L\\\frac{1}{3}\int\frac{1}{\sqrt{1-9x^{2}}}dx\)

Now, let \(\displaystyle \L\\x=\frac{1}{3}sin(9u), \;\ \frac{1}{3}dx=cos(9u)\)

This will simplify down to a very easy integral.