00:01
To evaluate this indefinite integral, the first thing you have to do is to apply substitution.
00:05
In here, we want to let u equal to ax.
00:09
Then, du is equal to a -d -x.
00:12
That means 1 over a -d -u is equal to d -x.
00:17
And then we can substitute it here.
00:19
We get negative 2 integral of 1 over a over u squared minus b squared raised to three over two and then you have d u or that's the same as negative 2 over a integral from or of 1 over u squared minus b squared raise to 3 halves d u and then we do trigonometric substitution and here we want to set u equal to b secant theta so that d u is equal to b secant theta tangent theta d theta so from here we have negative 2 over a integral of 1 over u squared that'll be b squared secan squared theta minus b squared raise to 3 over 2 and and then d is just b, secan, theta, tangent, theta, d theta.
01:30
And then we simplify the integral, we have negative 2 over a, integral of b secantheta, tangent theta, all over b squared raised to 3 over 2.
01:46
This time, second squared minus 1, raise to 3 over 2, and then d theta.
01:54
But since sequence squared theta minus 1 is just tangent squared theta, then from here we have negative 2 over a integral of secan theta, tangent theta, over b squared times tangent cubed theta, d theta.
02:20
And then this is just negative 2 over a, b squared, times secant theta, over tangent squared theta d theta.
02:31
And then you rewrite this further into negative 2 over a, b squared, integral of 1 over cosine theta times cosine squared theta over sine squared theta d theta.
02:46
That'll cancel out the cosines.
02:48
We have cosine here.
02:49
So from here we should have negative 2 over a, b squared, integral of cosine theta over sine d -theta, d -theta...