1. ## diff. trig integral

$\int tan^3(2t)sec^3(2t)dt = \int tan^2(2t)sec^2(2t)sec(2t)tan(2t)dt
$

$u = sec2t$
$du = sec(2t)tan(2t)dt$

$\int (tan^2(2t)u^2)du$

Am I going in the right direction?

2. Originally Posted by saiyanmx89
$\int tan^3(2t)sec^3(2t)dt = \int tan^2(2t)sec^2(2t)sec(2t)tan(2t)dt
$
Try this:

$\int\tan^32t\sec^32t\,dt$

$=\int\tan^22t\tan2t\sec^32t\,dt$

$=\int\left(\sec^22t-1\right)\tan2t\sec^32t\,dt$

$=\int\tan2t\sec^52t\,dt-\int\tan2t\sec^32t\,dt$

$=\int(\sec2t\tan2t)\sec^42t\,dt-\int(\sec2t\tan2t)\sec^22t\,dt$

Now do $u=\sec2t.$

3. You are going in the right direction, only


\begin{aligned}
u&=\sec 2t \\
du&=2 \sec 2t \tan 2t\,dt.
\end{aligned}

Now you can use the fact that $\tan^2 t+1=\sec^2 t$.

4. $\int \tan^3(2t)\sec^3(2t) \, dt$

$\int \tan^2(2t)\sec^2(2t)\sec(2t)\tan(2t) \, dt$

$\int [\sec^2(2t) - 1]\sec^2(2t)\sec(2t)\tan(2t) \, dt$

$\int [\sec^4(2t) - \sec^2(2t)]\sec(2t)\tan(2t) \, dt$

$u = \sec(2t)$

$du = 2\sec(2t)\tan(2t) \, dt$

$\frac{1}{2} \int [\sec^4(2t) - \sec^2(2t)] \cdot 2\sec(2t)\tan(2t) \, dt$

$\frac{1}{2} \int u^4 - u^2 \, du$

5. Originally Posted by Reckoner
Try this:

$\int\tan^32t\sec^32t\,dt$

$=\int\tan^22t\tan2t\sec^32t\,dt$

$=\int\left(\sec^22t-1\right)\tan2t\sec^32t\,dt$

$=\int\tan2t\sec^52t\,dt-\int\tan2t\sec^32t\,dt$

$=\int(\sec2t\tan2t)\sec^42t\,dt-\int(\sec2t\tan2t)\sec^22t\,dt$

Now do $u=\sec2t.$
what would du =?

6. Originally Posted by saiyanmx89
what would du =?
You should learn and know the derivatives of all six of the basic trigonometric functions.

$\frac d{dx}[\sec x]=\sec x\tan x$