$\int\sin^2(x) \, dx$ - Which one is right? If I enter $\int\sin^2(x)\,dx$ into an integral calculator it returns:
$$\int\sin^2(x)\,dx=-\frac{\sin(2x)-2x}{4}+c$$
If I calculate it with Angular multiples I come to:
$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{\sin(2x)-2x} 4 +c$$
Here is my calculation
If I calculate it with partial integration and substitution I come to:
$$\int\sin^2(x)\,dx=-\frac{\sin(x)\cos(x)+x} 2 +c$$
Here is my second calculation
If I plot all of the functions they look also different...
So which one is right? Or maybe are all of them right due to the "+c"?
EDIT:
I have added my calculations, maybe you can help me to spot the wrong signs - thanks!

With angular multiples,
$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{x}{2} - \frac{\sin(2x)}4 + c$
Now $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$\frac{x}{2} - \frac{2\sin(x)\cos(x)}4 + c$
$\frac{x}{2} - \frac{\sin(x)\cos(x)}2 + c$
$\frac{- \sin(x)\cos(x) + x}2 + c$
That is with partial integration and substitution.
Again $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$ \frac{- \sin(2x)}4 + \frac{x}{2} + c$
$ - \frac{\sin(2x) - 2x}4 + c$
That is with integral calculator.
So all are same. I think you have either typing mistake in your third solution or mistake in solving.
