1. ## another 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.

2. Originally Posted by davecs77
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

3. Originally Posted by topsquark
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
Why wouldnt you do it my way..by breaking both tan^4(x) and sec^4(x)? Would that just make things harder?..so a good strategy is to break one piece of it up? Thank you!

4. Originally Posted by davecs77
Why wouldnt you do it my way..by breaking both tan^4(x) and sec^4(x)? Would that just make things harder?..so a good strategy is to break one piece of it up? Thank you!
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?

5. Originally Posted by topsquark
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?

OOPS i meant integral of (sec^2(x) - 1)^2 + integral of (tan^2(x) + 1)^2

Thanks again..I was able to get the right answer now ;-)