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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!

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    You've got a sign error or two (e.g., an antiderivative of $dx/2$ is $x/2$, not $-x/2$, and an antiderivative of $\cos$ is $\sin$, not $-\sin$ :), but that aside, the two forms _are_ equivalent, thanks to the double-angle formula $\sin(2\theta) = 2\sin\theta \cos\theta$. – 2017-01-16
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    $\dfrac{\sin x\cos x} 2$ is the same thing as $\dfrac{\sin(2x)} 4,$ by a standard trigonometric identity. – 2017-01-16
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    I added my calculations, maybe you can help me spottin' the sign errors, because I couldn't find them... – 2017-01-16
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    @Gykonik: check what you missed in the second. – 2017-01-16
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    in the second what? :D – 2017-01-16
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    Possible duplicate of [What is the solution of $\int \sin(x)^2 dx$](https://math.stackexchange.com/questions/960465/what-is-the-solution-of-int-sinx2-dx) – 2017-11-11
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    Not a duplicate. The OP is not asking what the answer is - it is given right at the top. The OP is asking if the other forms are acceptable (no), or what is wrong with them (sign errors in the integration). The other thread does not address this. – 2017-11-11

3 Answers 3

1

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.

4

$$-\frac{\sin(2x)-2x}{4}+c=-\frac{2\sin(x)\cos (x)-2x}{4}+c=-\frac{\sin(x)\cos (x)-x}{2}+c$$

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:

$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{1}{2}\int(1-\cos(2x))dx=$$

$$\frac{1}{2}\left(\int(1)dx-\int\cos(2x)dx\right)=\frac{2x-\sin(2x)} 4 +c$$

4

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...

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    I added my calculations – 2017-01-16