How would you do this
integral of tan^4(x)sec^4(x)dx
I know this breakups into integral of ((sec^2(x))^2 + integral of ((tan^2(x))^2
I am not so sure where to go from here.
Thanks.
You know that $\displaystyle (tan(x))^{\prime} = sec^2(x)$, so
$\displaystyle \int tan^4(x) sec^4(x) dx = \int tan^4(x) sec^2(x) sec^2(x) dx$
What to do with the extra $\displaystyle sec^2(x)$? Well
$\displaystyle tan^2(x) + 1 = sec^2(x)$, so
$\displaystyle \int tan^4(x) sec^4(x) dx = \int tan^4(x) sec^2(x) sec^2(x) dx = \int tan^4(x) (tan^2(x) + 1) sec^2(x) dx $
Now sub in $\displaystyle y = tan(x)$ giving:
$\displaystyle \int tan^4(x) sec^4(x) dx = \int tan^4(x) (tan^2(x) + 1) sec^2(x) dx = \int y^4(y^2 + 1) dy$
You finish it from here.
-Dan
Because that's the way I'd do the integral and, honestly, I don't understand how
$\displaystyle \int tan^4(x) sec^4(x) dx = \int (sec^2(x))^2dx + \int (tan^2(x))^2 dx$. It just doesn't look right to me, though I admit I haven't spent any time to disprove it.
-Dan
Edit: The two expressions $\displaystyle tan^4(x) sec^4(x)$ and $\displaystyle sec^4(x) + tan^4(x)$ aren't the same. Where did you get this from?