Evaluate $\displaystyle \int\frac{{1}}{{{\arcsin{{\left({6}{x}\right)}}}\sqrt{{{1}-{36}{x}^{{2}}}}}}{\left.{d}{x}\right.}$ for $\displaystyle {0}<{x}<\frac{{1}}{{6}}$.

First make the substitution $\displaystyle {u}=$Preview Question 6 Part 1 of 4  

Then $\displaystyle \int\frac{{1}}{{{\arcsin{{\left({6}{x}\right)}}}\sqrt{{{1}-{36}{x}^{{2}}}}}}{\left.{d}{x}\right.}=\int$ Preview Question 6 Part 2 of 4   du

Now integrate with respect to $\displaystyle {u}$ to get + C   (in terms of $\displaystyle {u}$)  Preview Question 6 Part 3 of 4  

So $\displaystyle \int\frac{{1}}{{{\arcsin{{\left({6}{x}\right)}}}\sqrt{{{1}-{36}{x}^{{2}}}}}}{\left.{d}{x}\right.}$ = + C  Preview Question 6 Part 4 of 4