How do I evaluate the integral $\int\sin^3x dx$? $$\int \sin^3(x)dx$$
Using integration by parts:
$$u=\sin^3(x)$$
$$u'=3\sin^2(x)\cos(x)$$ 
$$v'=1$$
$$v=x$$
Gives:
$$\int \sin^3(x)dx= \sin^3(x)\cdot x  - 3 \int \sin^2(x)\cdot cos(x)dx$$
Using:
$$t=\sin(x)$$
$$\frac{dt}{dx}=\cos(x)$$
$$dt=\cos(x)dx$$ 
Results in:
$$=\sin^3(x)\cdot x - \sin^3(x) +C$$
I used integration by parts method
 A: By integration by parts you should have
$$\int \sin^3(x)dx= \sin^3(x)\cdot x  - 3 \int x\sin^2(x)\cdot \cos(x)dx.$$
The integral can be done in this way:
$$\int \sin^3(x)\, dx=\int (1-\cos^2(x))\sin(x)\, dx=\int \sin(x) dx+\int \cos^2(x)D(\cos(x))\, dx.$$
Can you take it from here?
A: If you use $u=\sin^3x$ and $v=1$, you get into big troubles, because the next integral you need to compute is
$$
\int 3x\sin^2x\cos x\,dx
$$
which is not at all easier than the one you started from.
If you really want to do it by parts, consider $u=\sin^2x$ and $v=\sin x$, so you get
\begin{align}
I&=\int\sin^3x\,dx \\
&=-\sin^2x\cos x+\int2\sin x\cos^2x\,dx \\
&=-\sin^2x\cos x+2\int\sin x\,dx-2\int\sin^3x\,dx \\
&=-\sin^2x\cos x-2\cos x-2I
\end{align}
that gives
$$
I=\frac{1}{3}(-\sin^2x\cos x-2\cos x)+c=\frac{1}{3}(\cos^3x-3\cos x)
$$
Much easier is to consider
$$
\int\sin^3x\,dx=\int(1-\cos^2x)\sin x\,dx\underset{u=\cos x}{=}\int(u^2-1)\,du
$$
which works alike for every odd power of $\sin x$.
A: Use $$\sin^3x=\frac{3\sin{x}-\sin3x}{4},$$
but the Robert Z's and the  Jack D'Aurizio's way is more nicer.
In your solution you need to write $x$ inside the last integral.
This is your mistake.
Because $\int{v'}dx=\int1dx=x+C$. 
A: $$\int\sin^3x\,\mathrm{d}x=\int\sin^2x\,\mathrm{d}(\sin x)\ldots$$
using power-reducation formula you can continue without by parts
remember that $$\sin^2x=\frac{1-\cos2x}{2}$$
