00:01
Hi, from the question, given that, consider the given integral 19 square root of x squared minus 4 divided by x to the power of 4 dx.
00:17
So in part a, we need to find which trigonometry substitution is correct for this integral.
00:23
So let x is equal to 2 secen theta.
00:28
So the above integral became integral 19 square root of, for x squared, 2 sicken theta, the old squared minus 4 divided by 2 second theta, the old power 4.
00:49
So from this, when you simplify it for dx, so we obtain 2 times of second theta, tan, theta, d theta.
01:01
So for dx substitute, two times of second theta, tan theta, d theta.
01:10
So this will become integral 19 times of square root of 2, 2 squared times.
01:21
2x2 x2x2 .2.
01:25
Divided by 2 power 4.
01:29
2 power 4 is 2 2 2 .4 4 2s are 8 2s are 16.
01:33
2.
01:35
2 .2.
01:36
2 .2.
01:41
2 2 .2.
01:41
2 2 2 2 2 2 .2.
01:43
10 theta d theta...