$\int\sin^2(x) \, dx$ - 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!
 A: $$-\frac{\sin(2x)-2x}{4}+c=-\frac{2\sin(x)\cos (x)-2x}{4}+c=-\frac{\sin(x)\cos (x)-x}{2}+c$$
PS: You are missing a signal in the $\color{red}{second}$ and $\color{red}{third}$ equalities.
The $\color{green}{first}$ is correct.
For the second:
$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{1}{2}\int(1-\cos(2x))dx=$$
$$\frac{1}{2}\left(\int(1)dx-\int\cos(2x)dx\right)=\frac{2x-\sin(2x)} 4 +c$$
A: Classic question.
Both answers are right! (The second one should have $\sin(2x)$, not $\sin 2$)
$$\frac{-\sin(2x)+2x}{4}+C=\frac{-2\sin(x)\cos(x)+2x}{4}+C=\frac{-\sin(x)\cos(x)+x}{2}.$$
Edit: You're missing a sign somewhere...
A: 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.
