Different answers for integral of $\sin^3x$ I was calculating an integral that I thought it'd be easy however it turned out very weird. I couldn't figure out so I'm looking for some help.
$$\int_{-\pi/2}^{\pi/2} \int_{2\cos\theta}^{2}2r{\sqrt{4-r^2}} \,dr\,d\vartheta$$
Then, I solved the inner integral with respect to r. I got the following by simplifying
$\int {2\over3}(4-(2\cos\theta)^2)^{3\over2} $
$\int {2\over3}(4-4\cos^2\theta)^{3\over2} $
$\int {2\over3}(4(1-\cos^2\theta))^{3\over2} $
$\int {2\over3}(4\sin^2\theta)^{3\over2} $
${16\over3}\int(\sin^3\theta) $
Then, with simple substitution, I got $\int(\sin^3\theta) d\theta $ = $1/3\cos^3\theta-\cos\theta+C$. When I plug in my end points $\pi/2$  and  $-\pi/2$, I get zero. However, it can't be zero because it's a legit solid piece. So, I tried to solve with wolphram alpha for $\sin^3\theta$ and my integrand before simplification, I got followings:


I also found this one for the integration of $\sin^3x$

 vs

So, I'm super confused how come $\sin^3x$ got two different answers???They seem the same but how come they are completely different functions? Where do I make mistake to solve the original problem? 
thank you
 A: There is no contradiction ...
If you use the substitution $y=\cos(x)$, you find that the primitives of $\sin^3$ are given by $x\mapsto\frac{1}{3}\cos^3(x)-\cos(x)+C$
On the other side, if you linearize first, you get $\sin^3(x)=\frac{1}{4}(3\sin(x)-\sin(3x))$, and hence the primitives of $\sin^3$ are given by $x\mapsto\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$
So, apparently, we get two different families of functions ... but only apparently because linearization of $\cos^3(x)$ leads to the formula :
$\cos^3(x)=\frac{1}{4}(\cos(3x)+3\cos(x))$
and so :
$\frac{1}{3}\cos^3(x)-\cos(x)=\frac{1}{12}(\cos(3x)+3\cos(x))-\cos(x)=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)$
A: $\int\sin^3xdx=\int(1-\cos^2x)\sin xdx=\int(\cos^2x-1)d(\cos x)=\frac{\cos^3x}{3}-\cos x+C$
A: There is actually no mistake. They are exactly the same since $\cos(3x)\equiv4\cos^3 x-3\cos x$. We demonstrate this as follows:
$$\cos(3x)=\cos(2x+x)$$
Using the double angle formula:
$$\cos(3x)=\cos(2x)\cdot\cos x-\sin(2x)\cdot \sin x$$
Using $\cos(2x)\equiv 2\cos^2 x-1$ and $\sin(2x)\equiv 2\sin x\cos x$, we obtain:
$$\cos(3x)=2\cos^3 x-\cos x-2\sin^2 x\cos x$$
Using the identity $\sin^2 x\equiv 1-\cos^2 x$, we obtain the correct identity:
$$\cos(3x)=2\cos^3 x-\cos x-2(1-\cos^2 x)\cos x$$
$$\boxed{\cos(3x)\equiv 4\cos^3 x-3\cos x}$$
Substituting this on your first solution gives the second solution.
