Am I taking the antiderivative of |x| correctly?

^{Edit: forgot +C, sorry}

So I remember seeing this video by Blackpenredpen on the Youtube homepage around 3 months ago where he was showing failed attempts of him trying to figure out how to get$$\int\sqrt{\sin^2(x)}dx=\int|\sin(x)|dx=-\cot(x)|\sin(x)|+c$$without piecewise integration.

While I do have my own attempts at this that I've done over the course of 2 months (by attempting IBP or using $u$-substitution), I think these attempts have failed because I don't really understand how to even evaluate$$\int|x|dx$$So I think that if I can integrate this, I can integrate$$\int|\sin(x)|dx$$using a similar method. Here is my attempt at doing so:

We have that$$\int|x|dx=\int1\cdot|x|dx$$Now, we can rewrite this as$$\int\left(\frac d{dx}x\right)|x|dx$$and now we can use IBP.

Here's a brief overview of IBP for anyone who doesn't remember it:

IBP, which stands for Integration By Parts, is a method of integration that states that for functions f(x) and g(x),$$\int f(x)dg(x)=f(x)g(x)-\int g(x)df(x)$$^{(here $dg(x)$ and $df(x)$ means $\frac d{dx}g(x)$ and $\frac d{dx}f(x)$ respectively. I understand the notation is disliked a lot)}
So what I think it is trying to show here in this context is that if we can rewrite our integral as a product of a 2 functions, and we know the antiderivative of one of them, we can use that to find the antiderivative of the other function.

So now we can use IBP on our integral$$I=\int\left(\frac d{dx}x\right)|x|dx$$using$$f(x)=|x|$$$$g(x)=x$$to get$$I=\int\left(\frac d{dx}x\right)|x|=x|x|-\int x\cdot\left(\frac{|x|}x\right)dx$$$$=x|x|-\int|x|dx$$$$I=x|x|-\int|x|dx\implies I=x|x|-I$$$$\implies 2I=x|x|\implies I=\int|x|dx=\frac{x|x|}2$$So we have that the antiderivative of $|x|$ is $\frac{x|x|}2+C$.

However, my question is: Am I taking the antiderivative of |x| correctly, or how would I take it?

posted 11 days ago

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11d ago

CrSb0001‭

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