00:01
For this problem we want to evaluate the integral of arc tangent of x raised a power of 6 d x using power series.
00:08
So first we recall the power series representation for arc tangent of x is equal to summation from n equals 0 to infinity of negative 1 raise of power of n times x raise 2n plus 1 over 2 n plus 1.
00:29
So if x is instead x raised a power of 6, then our arc tangent of x raise a power of 6, this is equal to summation from n equals 0 to infinity of negative 1 raise a power of n times x raise a power of 6, this raise 2n plus 1, all over 2n plus 1.
01:01
Simplifying, we have summation from n equals 0 to infinity of negative 1 raise of power of n times x raised to 12 n plus 6 all over 2n plus 1.
01:17
So if we take the antiderivative of arc tangent of x raised a power of 6 dx, that's taking the integral of the summation from n.
01:31
Equal 0 to infinity of negative 1 ratio power of n times x raise to 12 n plus 6 all over to 1 plus 1 d x.
01:43
And then you have to expand this.
01:44
We have integral of when n is 0, that's x to 6.
01:52
And then when n is 1, that's plus negative 1 times x raised to 12 plus 6, that's 18.
02:02
Over 2 plus 1, that's 3.
02:07
And then plus when x is 2 or when n is 2, we have 1 times x raise to 24 plus 6, that's 30, all over 2 times 2 plus 1, that's 5, and so on.
02:26
And then dx.
02:28
So we get integral of x to 6 minus x raise to a power of 18.
02:36
Over 3 plus x raised to 30 over 5 and so on.
02:43
This will be x raised the power of 7 over 7 minus x raised to 19 over 3 times 19 plus x raised to 30 plus 1 that's 31 over 5 times 31 and so on plus c or that summation from n equals 0 to infinity of negative 1 raised a power of n since it's alternating, times x raised 2.
03:20
If you notice, this is just 12n plus 6 plus 1, or 12n plus 7 all over.
03:29
The 2n plus 1 is still the same.
03:33
So 2n plus 1, but we will have to multiply this by 12n.
03:39
Plus 7 and then plus c...