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.

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- Jul 26th 2007, 06:19 PMdavecs77another trig sub.
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. - Jul 26th 2007, 06:23 PMtopsquark
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 - Jul 26th 2007, 06:30 PMdavecs77
- Jul 26th 2007, 06:35 PMtopsquark
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? - Jul 26th 2007, 06:50 PMdavecs77