# integrate $\int \tan^3(2t)\sec^3(2t) \, dt$

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

First try:$$\int \tan^3(2t)\sec^3(2t) \, dt=\int (\tan(2t)\sec(2t))^3 \, dt=\int \left(\frac{\sin(2t)}{\cos(2t)}\,\frac{1}{\cos(2t)}\right)^3 \, dt$$

$u=\cos2t$

$du=-2\sin(2t) \, dt$

$$-\frac{1}{2}\int \left(\frac{du}{u^2}\right)^3=\frac{1}{2} \cdot \frac{u}{5}^{-5}=\frac{1}{10\cos2t}+C$$

Second Try:

$$\int \tan^3(2t)\sec^3(2t) \, dt=\int \tan^2(2t)\sec^2(2t)\tan(2t)\sec(2t)dt=\int\sec^2(2t)(1-\sec^2(2t))\tan(2t)\sec(2t)$$

$u=sec(2t)$

$du=2sec(2t)tan(2t)$

$$\frac{1}{2}\int u^2(1-u^2)du=\frac{1}{2}\int u^2-u^4=\frac{u^3}{6}-\frac{u^5}{10}=\frac{sec^{3}(2t)}{6}-\frac{sec^{5}(2t)}{10}+c$$ Where have I got it wrong?

• Start by simplifying your integral by letting $2 t \to x$ and $d t = d x/2$. And it makes no sense to integrate over $(du)^3$. That's your problem. – David G. Stork Jan 9 '16 at 22:10
• @DavidG.Stork I need to find integrate according to $dt$ so it is just like $2x$ according to $dx$ – gbox Jan 9 '16 at 22:13
• I cannot see why you "need" to integrate according to $dt$. After all, you make a substitution to $du$. Why not first $dx$? – David G. Stork Jan 9 '16 at 22:14
• @DavidG.Stork ok so we can replace $dt$ in $dx$ – gbox Jan 9 '16 at 22:17

You should not get something like $\left(du \right)^3$ in the espression, it does not make sense. Indeed if you use $du=-2\sin(2t) \, dt$ where $u = \cos 2t$, you get $\sin^2 2t = 1- \cos^2 2t = 1 - u^2$. Put everything into the expression (i.e., express everything in terms of $u$), you get
$$\int \left(\frac{\sqrt{1-u^2}}{u^2}\right)^3 \frac{du}{-2 \sqrt{1-u^2}} = -\frac 12 \int \frac{1-u^2}{u^6} du.$$
• @gbox : Looks good, but it should be $u^2 -1$ instead of $1-u^2$, and you are missing some $du$'s. – user99914 Jan 9 '16 at 22:38
HINT: show that your integrand is equal to $$8\,{\frac { \left( \sin \left( t \right) \right) ^{3} \left( \cos \left( t \right) \right) ^{3}}{ \left( 2\, \left( \cos \left( t \right) \right) ^{2}-1 \right) ^{6}}}$$