Determine ∫tan8x sec4 x dx

To integrate tan⁡(8x)sec⁡4(x)tan(8x)sec4(x) with respect to xx, we can use the substitution method. Let's denote u=tan⁡(x)u=tan(x). Then du=sec⁡2(x) dxdu=sec2(x)dx.

Now, we need to express everything in terms of uu. First, we express tan⁡(8x)tan(8x) in terms of uu:

tan⁡(8x)=sin⁡(8x)cos⁡(8x)=sin⁡(8x)cos⁡4(x)cos⁡(8x)cos⁡4(x)=sin⁡(8x)cos⁡5(x)=2sin⁡(8x)(1+cos⁡(2x))5=2sin⁡(8x)(1+u2)5tan(8x)=cos(8x)sin(8x)=cos4(x)cos(8x)cos4(x)sin(8x)=cos5(x)sin(8x)=(1+cos(2x))52sin(8x)=(1+u2)52sin(8x)

Now, we have u=tan⁡(x)u=tan(x) and du=sec⁡2(x) dxdu=sec2(x)dx. Also, tan⁡(8x)=2sin⁡(8x)(1+u2)5tan(8x)=(1+u2)52sin(8x). So, our integral becomes:

∫2sin⁡(8x)(1+u2)5⋅1sec⁡2(x) du∫(1+u2)52sin(8x)⋅sec2(x)1du

=2∫sin⁡(8x)(1+u2)5 du=2∫(1+u2)5sin(8x)du

Now, we can use a reduction formula to integrate sin⁡(8x)(1+u2)5(1+u2)5sin(8x). Let's denote I(n)I(n) as the integral:

I(n)=∫sin⁡(8x)(1+u2)n duI(n)=∫(1+u2)nsin(8x)du

Then, we have:

I(n)=−1(n−1)(1+u2)n−1+u2(n−1)(n−3)(1+u2)n−3+1(n−1)(n−3)I(n−2)I(n)=−(n−1)(1+u2)n−11+(n−1)(n−3)(1+u2)n−3u2+(n−1)(n−3)1I(n−2)

Now, we apply this reduction formula to our integral:

I(5)=−14(1+u2)4+u24⋅2(1+u2)2+14⋅2I(3)I(5)=−4(1+u2)41+4⋅2(1+u2)2u2+4⋅21I(3)

I(5)=−14(1+u2)4+u28(1+u2)2+18I(3)I(5)=−4(1+u2)41+8(1+u2)2u2+81I(3)

Now, we need to find I(3)I(3):

I(3)=−12(1+u2)2+u22(1+u2)+12I(1)I(3)=−2(1+u2)21+2(1+u2)u2+21I(1)

I(3)=−12(1+u2)2+u22(1+u2)+12I(1)I(3)=−2(1+u2)21+2(1+u2)u2+21I(1)

Now, we need to find I(1)I(1). I(1)I(1) can be directly integrated:

I(1)=−cos⁡(8x)1+u2+arctan⁡(u)+CI(1)=−1+u2cos(8x)+arctan(u)+C

Now, we can substitute I(1)I(1) back into I(3)I(3) and I(3)I(3) back into I(5)I(5). Then, we substitute u=tan⁡(x)u=tan(x) back in terms of xx. We would end up with a long expression involving xx and tan⁡(x)tan(x).

However, please note that the reduction formula involves high-level algebraic manipulations, and it might be very complex to carry out by hand. If you have access to a symbolic computation software like Mathematica or Wolfram Alpha, you can use it to find the antiderivative easily.