Maximum Value of $\frac {\sin^3x\cos x}{\tan^2x+1}$

What is the maximum value of the following function? $$f(x)=\frac {\sin^3x\cos x}{\tan^2x+1}$$

I'm just not sure where I start. I have no requisite knowledge on finding the maximum values of any function.

And just a side question, is it possible to find the maximum/minimum of any function?

• Note $\tan^2 x + 1 = \sec^2 x$. No, it's not possible. Any just covers too much territory. – steven gregory Sep 28 '16 at 13:43
• $\tan^2 x+1=\sec^2 x$,Then it turns to be $\sin^3 x \cos^3 x$, I think you can do it. – gcy-rolle Sep 28 '16 at 13:45
• Or, what's the maximum of $s^3c^3$ (or maximum of $sc$) where $s^2+c^2=1$? (Silly me, just saw the answer, the bit that introduces $2x$!) – Heimdall Sep 28 '16 at 13:49

$$\frac { \sin ^{ 3 } x\cos x }{ \tan ^{ 2 } x+1 } =\frac { \sin ^{ 3 } x\cos x }{ \frac { 1 }{ \cos ^{ 2 }{ x } } } =\sin ^{ 3 }{ x\cos ^{ 3 }{ x } = } { \left( \frac { \sin { 2 } x }{ 2 } \right) }^{ 3 }=\frac { 1 }{ 8 } \sin ^{ 3 }{ 2x }$$ so max is ...