00:01
So in this problem, we're given the integral of the square root of 36 minus x squared over x dx.
00:16
And we're asked to calculate this using the substitution x equals 6 sine theta.
00:25
So we need to calculate what dx by d theta is.
00:30
This is simply 6 cosine theta, so dx is equal to 6 cos theta d theta.
00:45
Thus the integral becomes the square root of 36 minus 36 sine squared theta over 6 sine theta 6 cos theta d theta.
01:19
Those sixes will cancel, we get an extra factor of 6 from here, so we have 6 times the integral.
01:26
And then this becomes a 1 minus sine squared, which is cos squared square rooted.
01:31
So we have cos squared theta over sine theta d theta.
01:41
And the best way to tackle this is to re -expand this cos squared.
01:45
So we can write this as 6 times the integral of 1 minus sine squared theta over the sine of theta d theta, theta, which gives us 6 times the integral of 1 over sine, which is cosec, which gives us cosec of theta d theta minus the integral of sine theta d theta.
02:33
Now these are both standard integrals that you would look up.
02:38
So we have now six times the integral of cosec gives us minus the logarithm with the absolute value, using our absolute values here to be careful, of cosec theta plus cotangent of theta, and then our integral of sine becomes minus cosine.
03:22
So we have two minuses that cancel...