Question

What is the integral of `int tan^6(x)sec^6(x)`?

Answer 1

`tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C`

Explanation:

We will want to approach this so as to leave a `sec^2(x)` term hanging around in the integral. `(`This will act as `du` if we set `u=tan(x))`. If we convert the remaining `sec^4(x)` into functions of `tan(x)` through Pythagorean identities, then we can use `u` substitution.

`inttan^6(x)sec^6(x)dx`

`=inttan^6(x)sec^4(x)sec^2(x)dx`

`=inttan^6(x)(sec^2(x))^2sec^2(x)dx`

`=inttan^6(x)(1+tan^2(x))^2sec^2(x)dx`

`=inttan^6(x)(1+2tan^2(x)+tan^4(x))sec^2(x)dx`

`=int(tan^10(x)+2tan^8(x)+tan^6(x))sec^2(x)dx`

Now, let `u=tanx` which implies `du=sec^2(x)dx`.

Substituting, we see that

`=int(u^10+2u^8+u^6)du`

Integrating term by term, this gives

`=u^11/11+(2u^9)/9+u^7/7+C`

Since `u=tan(x)`:

`=tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C`