How Evaluate This Integral $\int x‎^{‎3‎} ‎‎\sin ‎2x ‎dx‎$? 
How can find the following indefinite integral?
  $$\int x‎^{‎3‎}  ‎‎\sin ‎2x ‎dx‎$$

Thanks For the Help!
 A: Let us consider the general case of $$I_n=\int x^n \sin(ax)\,dx$$ To make life simpler, change variable $ax=t$, $x=\frac t a$, $dx=\frac 1 a dt$. Replacing, this gives $$I_n=\frac 1{a^{n+1}}\int t^n \sin(t)\,dt=\frac 1{a^{n+1}}J_n$$ with $$J_n=\int t^n \sin(t)\,dt$$ Now, integrate by parts $u=t^n$, $dv=\sin(t)\,dt$, $du=nt^{n-1}\,dt$, $v=-\cos(t)$. So $$\int t^n \sin(t)\,dt=-t^n \cos(t) +n\int t^{n-1}\cos(t)\,dt$$ Integrate a second time by parts $u=t^{n-1}$, $dv=\cos(t)\,dt$, $du=(n-1)t^{n-2}\,dt$, $v=\sin(t)$. So, $$\int t^{n-1}\cos(t)\,dt=t^{n-1}\sin(t)-(n-1)\int t^{n-2} \sin(t)\,dt=t^{n-1}\sin(t)-(n-1)J_{n-2}$$ where you see appearing "almost" the same integral as at the begining with a lower power of $t$. Remember that $$J_0=-\cos (t)$$ $$J_1=\sin (t)-t \cos (t)$$
I am sure that you can take from here.
A: Hint: Integration by parts:
$\int_{x} x^{3}\sin 2x = \frac{-1}{2} \int_{x} x^{3} D\cos 2x = \frac{-1}{2}\cos (2x)x^{3} + \frac{3}{2} \int_{x} \cos (2x) x^{2} + \text{some constant}$; $\int_{x}\cos (2x)x^{2} = \frac{1}{2}\int_{x}x^{2}D\sin 2x = \frac{1}{2}\sin (2x)x^{2} - \frac{1}{2}\int_{x} \sin (2x)2x + \text{some constant}$.
A: For some particular integral problems which belong to 
$$A.T.E$$
$$Algebra\rightarrow Trigonometry\rightarrow Exponential$$
Take $u$ as first function and $v$ as second function satisfying $A\rightarrow T\rightarrow E$. Now 
$$\int uv=uv^{I}-u'v^{II}+u''v^{III}-u'''v^{IV}...$$
Where $u,u',u'',u''',u''''....$ are corresponding derivatives of function $u$. And $v^{I},v^{II},v^{III},v^{IV}...$ are corresponding integral of function $v$.  
