Question

Find `int tan^8x sec^4x dx`?

Answer 1

`int \ tan^8x \ sec^4x \ dx = 1/9tan^9x +1/11tan^11x + C`

Explanation:

We seek:

`int \ tan^8x \ sec^4x \ dx`

Using the identity:

`tan^2A+1-= sec^2A`

We can write the integral as:

`I = int \ tan^8x \ sec^2x \ sec^2x \ dx`
`\ \ = int \ tan^8x \ (1+tan^2x) \ sec^2x \ dx`
`\ \ = int \ (tan^8x +tan^10 x) \ sec^2x \ dx`

Then if we perform the substitution:

`u = tan x => (du)/dx = sec^2 x`

Then, we substitute into the internal to get:

`I = int \ (u^8 +u^10) \ du`

Which consists of standard functions, so we can readily integrate to get:

`I = 1/9u^9 +1/11u^11 + C`

Then we restore the substitution to get:

`I = 1/9tan^9x +1/11tan^11x + C`