Help in evaluating $\int \frac {t^4 \tan t}{2 + \cos t}~dt$

Can anybody help me in evaluating this indefinite integral? I can't possibly find a workable substitution: $$\int \dfrac {t^4 \tan t}{2 + \cos t}dt$$

I have already tried substituting $\tan t=\frac{\sin t}{\cos t}$ and it ended up as $\int \frac {t^4\frac {\sin t}{\cos t}}{2+\cos t}~dt \Rightarrow \int \frac {t^4\sin t}{2\cos t + \cos^2 t}~dt$. Substitution fails from here.

• Maybe you could try plugging in the definitions of $\cos (x)$ and $\tan (x)$ using $e$ or writing $\tan (t)=\frac{\sin (t)}{\cos (t)}$ to facilitate a substitution of some sort. Don't forget that there are various other identities for trig functions, too; write them nearby and play around with'm - see if inspiration strikes. That's how I would start. Nov 18 '13 at 14:10
• I would also try and approximate something that differentiates to become the integrand here, trying to refine my guesses iteratively as I go, but that's a long shot. Nov 18 '13 at 14:16
• Have you tried integration by parts? Nov 18 '13 at 14:20
• @Shaun ALready tried substituting $\tan t= \frac {\sin t}{\cos t}$ and it ended up as. $\int \frac {t^4\frac {\sin t}{\cos t}}{2+\cos t}~dt \Rightarrow \int \frac {t^4\sin t}{2\cos t + \cos^2 t}~dt$ Substitution fails from here. Nov 19 '13 at 0:27
• May 29 '14 at 16:46

You need to change it into:

$\int t^4 \tan(t) \times (2+\cos(t))^{-1} dt$

where:

$u = t^4 tan(t), \frac {du}{dx}=4t^5 \sec^2(t),$

$v=\int (2+cos(t))^{-1} dx,\frac{dv}{dx}=(2+cos(t))^{-1}$

then use this rule (Integration by parts):

$\int u \times \frac{dv}{dx} dx = u\times v-\int \frac{du}{dx} \times v$ $dx$